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leetcode-cn/originData/0Zeoeg.json
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"categoryTitle": "JavaScript",
|
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"boundTopicId": 2219915,
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"title": "Apply Transform Over Each Element in Array",
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"content": "<p>Given an integer array <code>arr</code> and a mapping function <code>fn</code>, return a new array with a transformation applied to each element.</p>\n\n<p>The returned array should be created such that <code>returnedArray[i] = fn(arr[i], i)</code>.</p>\n\n<p>Please solve it without the built-in <code>Array.map</code> method.</p>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> arr = [1,2,3], fn = function plusone(n) { return n + 1; }\n<strong>Output:</strong> [2,3,4]\n<strong>Explanation:</strong>\nconst newArray = map(arr, plusone); // [2,3,4]\nThe function increases each value in the array by one. \n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> arr = [1,2,3], fn = function plusI(n, i) { return n + i; }\n<strong>Output:</strong> [1,3,5]\n<strong>Explanation:</strong> The function increases each value by the index it resides in.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> arr = [10,20,30], fn = function constant() { return 42; }\n<strong>Output:</strong> [42,42,42]\n<strong>Explanation:</strong> The function always returns 42.\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>0 <= arr.length <= 1000</code></li>\n\t<li><code><font face=\"monospace\">-10<sup>9</sup> <= arr[i] <= 10<sup>9</sup></font></code></li>\n\t<li><font face=\"monospace\"><code>fn returns a number</code></font></li>\n</ul>\n",
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"translatedTitle": "转换数组中的每个元素",
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"Start by creating an array that will eventually be returned.",
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62
leetcode-cn/originData/array-prototype-last.json
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"boundTopicId": 2222272,
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"title": "Array Prototype Last",
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"content": "Write code that enhances all arrays such that you can call the <code>array.last()</code> method on any array and it will return the last element. If there are no elements in the array, it should return <code>-1</code>.\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [1,2,3]\n<strong>Output:</strong> 3\n<strong>Explanation:</strong> Calling nums.last() should return the last element: 3.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = []\n<strong>Output:</strong> -1\n<strong>Explanation:</strong> Because there are no elements, return -1.\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>0 <= arr.length <= 1000</code></li>\n\t<li><code>0 <= arr[i] <= 1000</code></li>\n</ul>\n",
|
||||
"translatedTitle": "数组原型对象的最后一个元素",
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||||
"translatedContent": "<p>请你编写一段代码实现一个数组方法，使任何数组都可以调用 <code>array.last()</code> 方法，这个方法将返回数组最后一个元素。如果数组中没有元素，则返回 <code>-1</code> 。</p>\n\n<p> </p>\n\n<p><strong>示例 1 ：</strong></p>\n\n<pre>\n<b>输入：</b>nums = [1,2,3]\n<b>输出：</b>3\n<b>解释</b>：调用 nums.last() 后返回最后一个元素： 3。\n</pre>\n\n<p><strong>示例 2 ：</strong></p>\n\n<pre>\n<b>输入：</b>nums = []\n<b>输出：</b>-1\n<strong>解释：</strong>因为此数组没有元素，所以应该返回 -1。\n</pre>\n\n<p> </p>\n\n<p><b>提示：</b></p>\n\n<ul>\n\t<li><code>0 <= arr.length <= 1000</code></li>\n\t<li><code>0 <= arr[i] <= 1000</code></li>\n</ul>\n",
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62
leetcode-cn/originData/array-reduce-transformation.json
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62
leetcode-cn/originData/array-reduce-transformation.json
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||||
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63
leetcode-cn/originData/cache-with-time-limit.json
Normal file
63
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"content": "<p>Write a class that allows getting and setting key-value pairs, however a <strong>time until expiration</strong> is associated with each key.</p>\n\n<p>The class has three public methods:</p>\n\n<p><code>set(key, value, duration)</code>: accepts an integer <code>key</code>, an integer <code>value</code>, and a <code>duration</code> in milliseconds. Once the <code>duration</code> has elapsed, the key should be inaccessible. The method should return <code>true</code> if the same un-expired key already exists and <code>false</code> otherwise. Both the value and duration should be overwritten if the key already exists.</p>\n\n<p><code>get(key)</code>: if an un-expired key exists, it should return the associated value. Otherwise it should return <code>-1</code>.</p>\n\n<p><code>count()</code>: returns the count of un-expired keys.</p>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> \n["TimeLimitedCache", "set", "get", "count", "get"]\n[[], [1, 42, 100], [1], [], [1]]\n[0, 0, 50, 50, 150]\n<strong>Output:</strong> [null, false, 42, 1, -1]\n<strong>Explanation:</strong>\nAt t=0, the cache is constructed.\nAt t=0, a key-value pair (1: 42) is added with a time limit of 100ms. The value doesn't exist so false is returned.\nAt t=50, key=1 is requested and the value of 42 is returned.\nAt t=50, count() is called and there is one active key in the cache.\nAt t=100, key=1 expires.\nAt t=150, get(1) is called but -1 is returned because the cache is empty.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> \n["TimeLimitedCache", "set", "set", "get", "get", "get", "count"]\n[[], [1, 42, 50], [1, 50, 100], [1], [1], [1], []]\n[0, 0, 40, 50, 120, 200, 250]\n<strong>Output:</strong> [null, false, true, 50, 50, -1]\n<strong>Explanation:</strong>\nAt t=0, the cache is constructed.\nAt t=0, a key-value pair (1: 42) is added with a time limit of 50ms. The value doesn't exist so false is returned.\nAt t=40, a key-value pair (1: 50) is added with a time limit of 100ms. A non-expired value already existed so true is returned and the old value was overwritten.\nAt t=50, get(1) is called which returned 50.\nAt t=120, get(1) is called which returned 50.\nAt t=140, key=1 expires.\nAt t=200, get(1) is called but the cache is empty so -1 is returned.\nAt t=250, count() returns 0 because the cache is empty.\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>0 <= key <= 10<sup>9</sup></code></li>\n\t<li><code>0 <= value <= 10<sup>9</sup></code></li>\n\t<li><code>0 <= duration <= 1000</code></li>\n\t<li><code>total method calls will not exceed 100</code></li>\n</ul>\n",
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"translatedContent": "<p>编写一个类，它允许获取和设置键-值对，并且每个键都有一个 <strong>过期时间</strong> 。</p>\n\n<p>该类有三个公共方法：</p>\n\n<p><code>set(key, value, duration)</code> ：接收参数为整型键 <code>key</code> 、整型值 <code>value</code> 和以毫秒为单位的持续时间 <code>duration</code> 。一旦 <code>duration</code> 到期后，这个键就无法访问。如果相同的未过期键已经存在，该方法将返回 <code>true</code> ，否则返回 <code>false</code> 。如果该键已经存在，则它的值和持续时间都应该被覆盖。</p>\n\n<p><code>get(key)</code> ：如果存在一个未过期的键，它应该返回这个键相关的值。否则返回 <code>-1</code> 。</p>\n\n<p><code>count()</code> ：返回未过期键的总数。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong> \n[\"TimeLimitedCache\", \"set\", \"get\", \"count\", \"get\"]\n[[], [1, 42, 100], [1], [], [1]]\n[0, 0, 50, 50, 150]\n<strong>输出：</strong> [null, false, 42, 1, -1]\n<strong>解释：</strong>\n在 t=0 时，缓存被构造。\n在 t=0 时，添加一个键值对 (1: 42) ，过期时间为 100ms 。因为该值不存在，因此返回false。\n在 t=50 时，请求 key=1 并返回值 42。\n在 t=50 时，调用 count() ，缓存中有一个未过期的键。\n在 t=100 时，key=1 到期。\n在 t=150 时，调用 get(1) ，返回 -1，因为缓存是空的。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>\n[\"TimeLimitedCache\", \"set\", \"set\", \"get\", \"get\", \"get\", \"count\"]\n[[], [1, 42, 50], [1, 50, 100], [1], [1], [1], []]\n[0, 0, 40, 50, 120, 200, 250]\n<strong>输出：</strong> [null, false, true, 50, 50, -1]\n<strong>解释：</strong>\n在 t=0 时，缓存被构造。\n在 t=0 时，添加一个键值对 (1: 42) ，过期时间为 50ms。因为该值不存在，因此返回false。\n当 t=40 时，添加一个键值对 (1: 50) ，过期时间为 100ms。因为一个未过期的键已经存在，返回 true 并覆盖这个键的旧值。\n在 t=50 时，调用 get(1) ，返回 50。\n在 t=120 时，调用 get(1) ，返回 50。\n在 t=140 时，key=1 过期。\n在 t=200 时，调用 get(1) ，但缓存为空，因此返回 -1。\n在 t=250 时，count() 返回0 ，因为缓存是空的，没有未过期的键。\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>0 <= key <= 10<sup>9</sup></code></li>\n\t<li><code>0 <= value <= 10<sup>9</sup></code></li>\n\t<li><code>0 <= duration <= 1000</code></li>\n\t<li><code>方法调用总数不会超过100</code></li>\n</ul>\n",
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||||
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||||
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|
||||
166
leetcode-cn/originData/calculate-delayed-arrival-time.json
Normal file
166
leetcode-cn/originData/calculate-delayed-arrival-time.json
Normal file
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||||
"title": "Check if Object Instance of Class",
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||||
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||||
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|
||||
"translatedTitle": "检查是否是类的对象实例",
|
||||
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||||
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||||
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|
||||
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|
||||
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|
||||
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|
||||
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||||
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||||
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63
leetcode-cn/originData/convert-object-to-json-string.json
Normal file
63
leetcode-cn/originData/convert-object-to-json-string.json
Normal file
@ -0,0 +1,63 @@
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||||
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||||
"content": "<p>Given an object, return a valid JSON string of that object. You may assume the object only inludes strings, integers, arrays, objects, booleans, and null. The returned string should not include extra spaces. The order of keys should be the same as the order returned by <code>Object.keys()</code>.</p>\n\n<p>Please solve it without using the built-in <code>JSON.stringify</code> method.</p>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> object = {"y":1,"x":2}\n<strong>Output:</strong> {"y":1,"x":2}\n<strong>Explanation:</strong> \nReturn the JSON representation.\nNote that the order of keys should be the same as the order returned by Object.keys().</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> object = {"a":"str","b":-12,"c":true,"d":null}\n<strong>Output:</strong> {"a":"str","b":-12,"c":true,"d":null}\n<strong>Explanation:</strong>\nThe primitives of JSON are strings, numbers, booleans, and null.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> object = {"key":{"a":1,"b":[{},null,"Hello"]}}\n<strong>Output:</strong> {"key":{"a":1,"b":[{},null,"Hello"]}}\n<strong>Explanation:</strong>\nObjects and arrays can include other objects and arrays.\n</pre>\n\n<p><strong class=\"example\">Example 4:</strong></p>\n\n<pre>\n<strong>Input:</strong> object = true\n<strong>Output:</strong> true\n<strong>Explanation:</strong>\nPrimitive types are valid inputs.</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>object includes strings, integers, booleans, arrays, objects, and null</code></li>\n\t<li><code>1 <= JSON.stringify(object).length <= 10<sup>5</sup></code></li>\n\t<li><code>maxNestingLevel <= 1000</code></li>\n\t<li><code>all strings will only contain alphanumeric characters</code></li>\n</ul>\n",
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||||
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|
||||
"translatedContent": "<p>现给定一个对象，返回该对象的有效 JSON 字符串。你可以假设这个对象只包括字符串、整数、数组、对象、布尔值和 null。返回的字符串不能包含额外的空格。键的返回顺序应该与 <code>Object.keys()</code> 的顺序相同。</p>\n\n<p>请你在不使用内置方法 <code>JSON.stringify</code> 的前提下解决这个问题。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<b>输入：</b>object = {\"y\":1,\"x\":2}\n<b>输出：</b>{\"y\":1,\"x\":2}\n<b>解释：</b>\n返回该对象的 JSON 表示形式。\n注意，键的返回顺序应该与 Object.keys() 的顺序相同。</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>object = {\"a\":\"str\",\"b\":-12,\"c\":true,\"d\":null}\n<b>输出：</b>{\"a\":\"str\",\"b\":-12,\"c\":true,\"d\":null}\n<strong>解释：</strong>\nJSON 的基本类型是字符串、数字型、布尔值和 null。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<b>输入：</b>object = {\"key\":{\"a\":1,\"b\":[{},null,\"Hello\"]}}\n<b>输出：</b>{\"key\":{\"a\":1,\"b\":[{},null,\"Hello\"]}}\n<b>解释：</b>\n对象和数组可以包括其他对象和数组。\n</pre>\n\n<p><strong>示例 4：</strong></p>\n\n<pre>\n<b>输入：</b>object = true\n<b>输出：</b>true\n<b>解释：</b>\n基本类型是有效的输入</pre>\n\n<p> </p>\n\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>对象包括字符串、整数、布尔值、数组、对象和 null</code></li>\n\t<li><code>1 <= JSON.stringify(object).length <= 10<sup>5</sup></code></li>\n\t<li><code>maxNestingLevel <= 1000</code></li>\n</ul>\n",
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||||
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||||
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||||
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62
leetcode-cn/originData/counter.json
Normal file
62
leetcode-cn/originData/counter.json
Normal file
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||||
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||||
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||||
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||||
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|
||||
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||||
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196
leetcode-cn/originData/cousins-in-binary-tree-ii.json
Normal file
196
leetcode-cn/originData/cousins-in-binary-tree-ii.json
Normal file
File diff suppressed because one or more lines are too long
62
leetcode-cn/originData/curry.json
Normal file
62
leetcode-cn/originData/curry.json
Normal file
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||||
"content": "<p>Given a function <code>fn</code>, return a <strong>curried</strong> version of that function.</p>\n\n<p>A <strong>curried</strong> function is a function that accepts fewer or an equal number of parameters as the original function and returns either another <strong>curried</strong> function or the same value the original function would have returned.</p>\n\n<p>In practical terms, if you called the original function like <code>sum(1,2,3)</code>, you would call the <strong>curried</strong> version like <code>csum(1)(2)(3)<font face=\"sans-serif, Arial, Verdana, Trebuchet MS\">, </font></code><code>csum(1)(2,3)</code>, <code>csum(1,2)(3)</code>, or <code>csum(1,2,3)</code>. All these methods of calling the <strong>curried</strong> function should return the same value as the original.</p>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> \nfn = function sum(a, b, c) { return a + b + c; }\ninputs = [[1],[2],[3]]\n<strong>Output:</strong> 6\n<strong>Explanation:</strong>\nThe code being executed is:\nconst curriedSum = curry(fn);\ncurriedSum(1)(2)(3) === 6;\ncurriedSum(1)(2)(3) should return the same value as sum(1, 2, 3).\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong>\nfn = function sum(a, b, c) { return a + b + c; }\ninputs = [[1,2],[3]]]\n<strong>Output:</strong> 6\n<strong>Explanation:</strong>\ncurriedSum(1, 2)(3) should return the same value as sum(1, 2, 3).</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong>\nfn = function sum(a, b, c) { return a + b + c; }\ninputs = [[],[],[1,2,3]]\n<strong>Output:</strong> 6\n<strong>Explanation:</strong>\nYou should be able to pass the parameters in any way, including all at once or none at all.\ncurriedSum()()(1, 2, 3) should return the same value as sum(1, 2, 3).\n</pre>\n\n<p><strong class=\"example\">Example 4:</strong></p>\n\n<pre>\n<strong>Input:</strong>\nfn = function life() { return 42; }\ninputs = [[]]\n<strong>Output:</strong> 42\n<strong>Explanation:</strong>\ncurrying a function that accepts zero parameters should effectively do nothing.\ncurriedLife() === 42\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 <= inputs.length <= 1000</code></li>\n\t<li><code>0 <= inputs[i][j] <= 10<sup>5</sup></code></li>\n\t<li><code>0 <= fn.length <= 1000</code></li>\n\t<li><code>inputs.flat().length == fn.length</code></li>\n\t<li><code>function parameters explicitly defined</code></li>\n</ul>\n",
|
||||
"translatedTitle": "柯里化",
|
||||
"translatedContent": "<p>请你编写一个函数，它接收一个其他的函数，并返回该函数的 <strong>柯里化 </strong>后的形式。</p>\n\n<p><strong>柯里化 </strong>函数的定义是接受与原函数相同数量或更少数量的参数，并返回另一个 <strong>柯里化</strong> 后的函数或与原函数相同的值。</p>\n\n<p>实际上，当你调用原函数，如 <code>sum(1,2,3)</code> 时，它将调用 <strong>柯里化</strong> 函数的某个形式，如 <code>csum(1)(2)(3)</code>， <code>csum(1)(2,3)</code>， <code>csum(1,2)(3)</code>，或 <code>csum(1,2,3)</code> 。所有调用 <strong>柯里化</strong> 函数的方法都应该返回与原始函数相同的值。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<b>输入：</b>\nfn = function sum(a, b, c) { return a + b + c; }\ninputs = [[1],[2],[3]]\n<b>输出：</b>6\n<strong>解释：</strong>\n执行的代码是：\nconst curriedSum = curry(fn);\ncurriedSum(1)(2)(3) === 6;\ncurriedSum(1)(2)(3) 应该返回像原函数 sum(1, 2, 3) 一样的值。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>\nfn = function sum(a, b, c) { return a + b + c; }\ninputs = [[1,2],[3]]]\n<b>输出：</b>6\n<strong>解释：</strong>\ncurriedSum(1, 2)(3) 应该返回像原函数 sum(1, 2, 3) 一样的值。</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>\nfn = function sum(a, b, c) { return a + b + c; }\ninputs = [[],[],[1,2,3]]\n<b>输出：</b>6\n<strong>解释：</strong>\n你应该能够以任何方式传递参数，包括一次性传递所有参数或完全不传递参数。\ncurriedSum()()(1, 2, 3) 应该返回像原函数 sum(1, 2, 3) 一样的值。\n</pre>\n\n<p><strong>示例 4：</strong></p>\n\n<pre>\n<strong>输入：</strong>\nfn = function life() { return 42; }\ninputs = [[]]\n<b>输出：</b>42\n<strong>解释：</strong>\n柯里化一个没有接收参数，没做有效操作的函数。\ncurriedLife() === 42\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 <= inputs.length <= 1000</code></li>\n\t<li><code>0 <= inputs[i][j] <= 10<sup>5</sup></code></li>\n\t<li><code>0 <= fn.length <= 1000</code></li>\n\t<li><code>inputs.flat().length == fn.length</code></li>\n\t<li><code>函数参数需要被显式定义</code></li>\n</ul>\n",
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||||
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62
leetcode-cn/originData/debounce.json
Normal file
62
leetcode-cn/originData/debounce.json
Normal file
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"content": "<p>Given a function <code>fn</code> and a time in milliseconds <code>t</code>, return a <strong>debounced</strong> version of that function.</p>\n\n<p>A <strong>debounced</strong> function is a function whose execution is delayed by <code>t</code> milliseconds and whose execution is cancelled if it is called again within that window of time. The debounced function should also recieve the passed parameters.</p>\n\n<p>For example, let's say <code>t = 50ms</code>, and the function was called at <code>30ms</code>, <code>60ms</code>, and <code>100ms</code>. The first 2 function calls would be cancelled, and the 3rd function call would be executed at <code>150ms</code>. If instead <code>t = 35ms</code>, The 1st call would be cancelled, the 2nd would be executed at <code>95ms</code>, and the 3rd would be executed at <code>135ms</code>.</p>\n\n<p><img alt=\"Debounce Schematic\" src=\"https://assets.leetcode.com/uploads/2023/04/08/screen-shot-2023-04-08-at-11048-pm.png\" style=\"width: 800px; height: 242px;\" /></p>\n\n<p>The above diagram shows how debounce will transform events. Each rectangle represents 100ms and the debounce time is 400ms. Each color represents a different set of inputs.</p>\n\n<p>Please solve it without using lodash's <code>_.debounce()</code> function.</p>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> \nt = 50\ncalls = [\n {"t": 50, inputs: [1]},\n {"t": 75, inputs: [2]}\n]\n<strong>Output:</strong> [{"t": 125, inputs: [2]}]\n<strong>Explanation:</strong>\nlet start = Date.now();\nfunction log(...inputs) { \n console.log([Date.now() - start, inputs ])\n}\nconst dlog = debounce(log, 50);\nsetTimeout(() => dlog(1), 50);\nsetTimeout(() => dlog(2), 75);\n\nThe 1st call is cancelled by the 2nd call because the 2nd call occurred before 100ms\nThe 2nd call is delayed by 50ms and executed at 125ms. The inputs were (2).\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> \nt = 20\ncalls = [\n {"t": 50, inputs: [1]},\n {"t": 100, inputs: [2]}\n]\n<strong>Output:</strong> [{"t": 70, inputs: [1]}, {"t": 120, inputs: [2]}]\n<strong>Explanation:</strong>\nThe 1st call is delayed until 70ms. The inputs were (1).\nThe 2nd call is delayed until 120ms. The inputs were (2).\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> \nt = 150\ncalls = [\n {"t": 50, inputs: [1, 2]},\n {"t": 300, inputs: [3, 4]},\n {"t": 300, inputs: [5, 6]}\n]\n<strong>Output:</strong> [{"t": 200, inputs: [1,2]}, {"t": 450, inputs: [5, 6]}]\n<strong>Explanation:</strong>\nThe 1st call is delayed by 150ms and ran at 200ms. The inputs were (1, 2).\nThe 2nd call is cancelled by the 3rd call\nThe 3rd call is delayed by 150ms and ran at 450ms. The inputs were (5, 6).\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>0 <= t <= 1000</code></li>\n\t<li><code>1 <= calls.length <= 10</code></li>\n\t<li><code>0 <= calls[i].t <= 1000</code></li>\n\t<li><code>0 <= calls[i].inputs.length <= 10</code></li>\n</ul>\n",
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"translatedTitle": "函数防抖",
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"translatedContent": "<p>请你编写一个函数，接收参数为另一个函数和一个以毫秒为单位的时间 <code>t</code> ，并返回该函数的 <b>函数防抖 </b>后的结果。</p>\n\n<p><b>函数防抖 </b>方法是一个函数，它的执行被延迟了 <code>t</code> 毫秒，如果在这个时间窗口内再次调用它，它的执行将被取消。你编写的防抖函数也应该接收传递的参数。</p>\n\n<p>例如，假设 <code>t = 50ms</code> ，函数分别在 <code>30ms</code> 、 <code>60ms</code> 和 <code>100ms</code> 时调用。前两个函数调用将被取消，第三个函数调用将在 <code>150ms</code> 执行。如果改为 <code>t = 35ms</code> ，则第一个调用将被取消，第二个调用将在 <code>95ms</code> 执行，第三个调用将在 <code>135ms</code> 执行。</p>\n\n<p><img alt=\"Debounce Schematic\" src=\"https://assets.leetcode.com/uploads/2023/04/08/screen-shot-2023-04-08-at-11048-pm.png\" style=\"width: 800px; height: 242px;\" /></p>\n\n<p>上图展示了了防抖函数是如何转换事件的。其中，每个矩形表示 100ms，反弹时间为 400ms。每种颜色代表一组不同的输入。</p>\n\n<p>请在不使用 lodash 的 <code>_.debounce()</code> 函数的前提下解决该问题。</p>\n\n<p> </p>\n\n<p><strong class=\"example\">示例 1：</strong></p>\n\n<pre>\n<b>输入：</b>\nt = 50\ncalls = [\n {\"t\": 50, inputs: [1]},\n {\"t\": 75, inputs: [2]}\n]\n<b>输出：</b>[{\"t\": 125, inputs: [2]}]\n<strong>解释：</strong>\nlet start = Date.now();\nfunction log(...inputs) { \n console.log([Date.now() - start, inputs ])\n}\nconst dlog = debounce(log, 50);\nsetTimeout(() => dlog(1), 50);\nsetTimeout(() => dlog(2), 75);\n\n第一次调用被第二次调用取消，因为第二次调用发生在 100ms 之前\n第二次调用延迟 50ms，在 125ms 执行。输入为 (2)。\n</pre>\n\n<p><strong class=\"example\">示例 2：</strong></p>\n\n<pre>\n<b>输入：</b>\nt = 20\ncalls = [\n {\"t\": 50, inputs: [1]},\n {\"t\": 100, inputs: [2]}\n]\n<b>输出：</b>[{\"t\": 70, inputs: [1]}, {\"t\": 120, inputs: [2]}]\n<strong>解释：</strong>\n第一次调用延迟到 70ms。输入为 (1)。\n第二次调用延迟到 120ms。输入为 (2)。\n</pre>\n\n<p><strong class=\"example\">示例 3：</strong></p>\n\n<pre>\n<b>输入：</b>\nt = 150\ncalls = [\n {\"t\": 50, inputs: [1, 2]},\n {\"t\": 300, inputs: [3, 4]},\n {\"t\": 300, inputs: [5, 6]}\n]\n<b>输出：</b>[{\"t\": 200, inputs: [1,2]}, {\"t\": 450, inputs: [5, 6]}]\n<strong>解释：</strong>\n第一次调用延迟了 150ms，运行时间为 200ms。输入为 (1, 2)。\n第二次调用被第三次调用取消\n第三次调用延迟了 150ms，运行时间为 450ms。输入为 (5, 6)。\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>0 <= t <= 1000</code></li>\n\t<li><code>1 <= calls.length <= 10</code></li>\n\t<li><code>0 <= calls[i].t <= 1000</code></li>\n\t<li><code>0 <= calls[i].inputs.length <= 10</code></li>\n</ul>\n",
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63
leetcode-cn/originData/design-cancellable-function.json
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63
leetcode-cn/originData/design-cancellable-function.json
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63
leetcode-cn/originData/filter-elements-from-array.json
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leetcode-cn/originData/filter-elements-from-array.json
Normal file
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"content": "<p>Given an integer array <code>arr</code> and a filtering function <code>fn</code>, return a new array with a fewer or equal number of elements.</p>\n\n<p>The returned array should only contain elements where <code>fn(arr[i], i)</code> evaluated to a truthy value.</p>\n\n<p>Please solve it without the built-in <code>Array.filter</code> method.</p>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> arr = [0,10,20,30], fn = function greaterThan10(n) { return n > 10; }\n<strong>Output:</strong> [20,30]\n<strong>Explanation:</strong>\nconst newArray = filter(arr, fn); // [20, 30]\nThe function filters out values that are not greater than 10</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> arr = [1,2,3], fn = function firstIndex(n, i) { return i === 0; }\n<strong>Output:</strong> [1]\n<strong>Explanation:</strong>\nfn can also accept the index of each element\nIn this case, the function removes elements not at index 0\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> arr = [-2,-1,0,1,2], fn = function plusOne(n) { return n + 1 }\n<strong>Output:</strong> [-2,0,1,2]\n<strong>Explanation:</strong>\nFalsey values such as 0 should be filtered out\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>0 <= arr.length <= 1000</code></li>\n\t<li><code><font face=\"monospace\">-10<sup>9</sup> <= arr[i] <= 10<sup>9</sup></font></code></li>\n</ul>\n",
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"translatedContent": "<p>请你编写一个函数，该函数接受一个整数数组参数 <code>arr</code> 和一个过滤函数 <code>fn</code>，并返回一个过滤后元素数量较少或元素数量相等的新数组。</p>\n\n<p>返回的数组应该只包含通过过滤函数 <code>fn(arr[i]， i)</code> 计算后为真值的元素。</p>\n\n<p>请你在不使用内置函数 <code>Array.filter</code> 的前提下解决该问题。</p>\n\n<p> </p>\n\n<p><strong class=\"example\">示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>arr = [0,10,20,30], fn = function greaterThan10(n) { return n > 10; }\n<b>输出：</b> [20,30]\n<b>解释：</b>\nconst newArray = filter(arr, fn); // [20, 30]\n过滤函数过滤掉不大于 10 的值</pre>\n\n<p><strong class=\"example\">示例 2：</strong></p>\n\n<pre>\n<b>输入：</b>arr = [1,2,3], fn = function firstIndex(n, i) { return i === 0; }\n<b>输出：</b>[1]\n<strong>解释：</strong>\n过滤函数 fn 也可以接受每个元素的索引\n在这种情况下，过滤函数删除索引不为 0 的元素\n</pre>\n\n<p><strong class=\"example\">示例 3：</strong></p>\n\n<pre>\n<b>输入：</b>arr = [-2,-1,0,1,2], fn = function plusOne(n) { return n + 1 }\n<b>输出：</b>[-2,0,1,2]\n<strong>解释：</strong>\n像 0 这样的假值应被过滤掉\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>0 <= arr.length <= 1000</code></li>\n\t<li><code><font face=\"monospace\">-10<sup>9</sup> <= arr[i] <= 10<sup>9</sup></font></code></li>\n</ul>\n",
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171
leetcode-cn/originData/find-the-maximum-divisibility-score.json
Normal file
171
leetcode-cn/originData/find-the-maximum-divisibility-score.json
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177
leetcode-cn/originData/find-the-width-of-columns-of-a-grid.json
Normal file
177
leetcode-cn/originData/find-the-width-of-columns-of-a-grid.json
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62
leetcode-cn/originData/function-composition.json
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62
leetcode-cn/originData/function-composition.json
Normal file
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"content": "<p>Given an array of functions <code>[f<span style=\"font-size: 10.8333px;\">1</span>, f<sub>2</sub>, f<sub>3</sub>, ..., f<sub>n</sub>]</code>, return a new function <code>fn</code> that is the <strong>function composition</strong> of the array of functions.</p>\n\n<p>The <strong>function composition</strong> of <code>[f(x), g(x), h(x)]</code> is <code>fn(x) = f(g(h(x)))</code>.</p>\n\n<p>The <strong>function composition</strong> of an empty list of functions is the <strong>identity function</strong> <code>f(x) = x</code>.</p>\n\n<p>You may assume each function in the array accepts one integer as input and returns one integer as output.</p>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> functions = [x => x + 1, x => x * x, x => 2 * x], x = 4\n<strong>Output:</strong> 65\n<strong>Explanation:</strong>\nEvaluating from right to left ...\nStarting with x = 4.\n2 * (4) = 8\n(8) * (8) = 64\n(64) + 1 = 65\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> functions = [x => 10 * x, x => 10 * x, x => 10 * x], x = 1\n<strong>Output:</strong> 1000\n<strong>Explanation:</strong>\nEvaluating from right to left ...\n10 * (1) = 10\n10 * (10) = 100\n10 * (100) = 1000\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> functions = [], x = 42\n<strong>Output:</strong> 42\n<strong>Explanation:</strong>\nThe composition of zero functions is the identity function</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code><font face=\"monospace\">-1000 <= x <= 1000</font></code></li>\n\t<li><code><font face=\"monospace\">0 <= functions.length <= 1000</font></code></li>\n\t<li><font face=\"monospace\"><code>all functions accept and return a single integer</code></font></li>\n</ul>\n",
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64
leetcode-cn/originData/generate-fibonacci-sequence.json
Normal file
64
leetcode-cn/originData/generate-fibonacci-sequence.json
Normal file
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||||
"translatedTitle": "生成斐波那契数列",
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||||
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63
leetcode-cn/originData/group-by.json
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Normal file
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"content": "<p>Write code that enhances all arrays such that you can call the <code>array.groupBy(fn)</code> method on any array and it will return a <strong>grouped</strong> version of the array.</p>\n\n<p>A <strong>grouped</strong> array is an object where each key is the output of <code>fn(arr[i])</code> and each value is an array containing all items in the original array with that key.</p>\n\n<p>The provided callback <code>fn</code> will accept an item in the array and return a string key.</p>\n\n<p>The order of each value list should be the order the items appear in the array. Any order of keys is acceptable.</p>\n\n<p>Please solve it without lodash's <code>_.groupBy</code> function.</p>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> \narray = [\n {"id":"1"},\n {"id":"1"},\n {"id":"2"}\n], \nfn = function (item) { \n return item.id; \n}\n<strong>Output:</strong> \n{ \n "1": [{"id": "1"}, {"id": "1"}], \n "2": [{"id": "2"}] \n}\n<strong>Explanation:</strong>\nOutput is from array.groupBy(fn).\nThe selector function gets the "id" out of each item in the array.\nThere are two objects with an "id" of 1. Both of those objects are put in the first array.\nThere is one object with an "id" of 2. That object is put in the second array.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> \narray = [\n [1, 2, 3],\n [1, 3, 5],\n [1, 5, 9]\n]\nfn = function (list) { \n return String(list[0]); \n}\n<strong>Output:</strong> \n{ \n "1": [[1, 2, 3], [1, 3, 5], [1, 5, 9]] \n}\n<strong>Explanation:</strong>\nThe array can be of any type. In this case, the selector function defines the key as being the first element in the array. \nAll the arrays have 1 as their first element so they are grouped together.\n{\n "1": [[1, 2, 3], [1, 3, 5], [1, 5, 9]]\n}\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> \narray = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]\nfn = function (n) { \n return String(n > 5);\n}\n<strong>Output:</strong>\n{\n "true": [6, 7, 8, 9, 10],\n "false": [1, 2, 3, 4, 5]\n}\n<strong>Explanation:</strong>\nThe selector function splits the array by whether each number is greater than 5.\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>0 <= array.length <= 10<sup>5</sup></code></li>\n\t<li><code>fn returns a string</code></li>\n</ul>\n",
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||||
"translatedTitle": "分组",
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||||
"translatedContent": "<p>请你编写一段可应用于所有数组的代码，使任何数组调用 <code>array. groupBy(fn)</code> 方法时，它返回对该数组 <strong>分组后</strong> 的结果。</p>\n\n<p>数组 <strong>分组</strong> 是一个对象，其中的每个键都是 <code>fn(arr[i])</code> 的输出的一个数组，该数组中含有原数组中具有该键的所有项。</p>\n\n<p>提供的回调函数 <code>fn</code> 将接受数组中的项并返回一个字符串类型的键。</p>\n\n<p>每个值列表的顺序应该与元素在数组中出现的顺序相同。任何顺序的键都是可以接受的。</p>\n\n<p>请在不使用 lodash 的 <code>_.groupBy</code> 函数的前提下解决这个问题。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<b>输入：</b>\narray = [\n {\"id\":\"1\"},\n {\"id\":\"1\"},\n {\"id\":\"2\"}\n], \nfn = function (item) { \n return item.id; \n}\n<b>输出：</b>\n{ \n \"1\": [{\"id\": \"1\"}, {\"id\": \"1\"}], \n \"2\": [{\"id\": \"2\"}] \n}\n<strong>解释：</strong>\n输出来自函数 array.groupBy(fn)。\n分组选择方法是从数组中的每个项中获取 \"id\" 。\n有两个 \"id\" 为 1 的对象。所以将这两个对象都放在第一个数组中。\n有一个 \"id\" 为 2 的对象。所以该对象被放到第二个数组中。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<b>输入：</b>\narray = [\n [1, 2, 3],\n [1, 3, 5],\n [1, 5, 9]\n]\nfn = function (list) { \n return String(list[0]); \n}\n<b>输出：</b>\n{ \n \"1\": [[1, 2, 3], [1, 3, 5], [1, 5, 9]] \n}\n<strong>解释：</strong>\n数组可以是任何类型的。在本例中，分组选择方法是将键定义为数组中的第一个元素。\n所有数组的第一个元素都是1，所以它们被组合在一起。\n{\n \"1\": [[1, 2, 3], [1, 3, 5], [1, 5, 9]]\n}\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<b>输出：</b>\narray = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]\nfn = function (n) { \n return String(n > 5);\n}\n<strong>输入：</strong>\n{\n \"true\": [6, 7, 8, 9, 10],\n \"false\": [1, 2, 3, 4, 5]\n}\n<strong>解释：</strong>\n分组选择方法是根据每个数字是否大于 5 来分割数组。\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>0 <= array.length <= 10<sup>5</sup></code></li>\n\t<li><code>fn 返回一个字符串</code></li>\n</ul>\n",
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161
leetcode-cn/originData/hqCnmP.json
Normal file
161
leetcode-cn/originData/hqCnmP.json
Normal file
File diff suppressed because one or more lines are too long
63
leetcode-cn/originData/json-deep-equal.json
Normal file
63
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Normal file
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64
leetcode-cn/originData/memoize-ii.json
Normal file
64
leetcode-cn/originData/memoize-ii.json
Normal file
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"content": "<p>Given a function <code>fn</code>, return a <strong>memoized</strong> version of that function.</p>\n\n<p>A <strong>memoized </strong>function is a function that will never be called twice with the same inputs. Instead it will return a cached value.</p>\n\n<p><code>fn</code> can be any function and there are no constraints on what type of values it accepts. Inputs are considered identical if they are <code>===</code> to each other.</p>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> \ngetInputs = () => [[2,2],[2,2],[1,2]]\nfn = function (a, b) { return a + b; }\n<strong>Output:</strong> [{"val":4,"calls":1},{"val":4,"calls":1},{"val":3,"calls":2}]\n<strong>Explanation:</strong>\nconst inputs = getInputs();\nconst memoized = memoize(fn);\nfor (const arr of inputs) {\n memoized(...arr);\n}\n\nFor the inputs of (2, 2): 2 + 2 = 4, and it required a call to fn().\nFor the inputs of (2, 2): 2 + 2 = 4, but those inputs were seen before so no call to fn() was required.\nFor the inputs of (1, 2): 1 + 2 = 3, and it required another call to fn() for a total of 2.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> \ngetInputs = () => [[{},{}],[{},{}],[{},{}]] \nfn = function (a, b) { return ({...a, ...b}); }\n<strong>Output:</strong> [{"val":{},"calls":1},{"val":{},"calls":2},{"val":{},"calls":3}]\n<strong>Explanation:</strong>\nMerging two empty objects will always result in an empty object. It may seem like there should only be 1 call to fn() because of cache-hits, however none of those objects are === to each other.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> \ngetInputs = () => { const o = {}; return [[o,o],[o,o],[o,o]]; }\nfn = function (a, b) { return ({...a, ...b}); }\n<strong>Output:</strong> [{"val":{},"calls":1},{"val":{},"calls":1},{"val":{},"calls":1}]\n<strong>Explanation:</strong>\nMerging two empty objects will always result in an empty object. The 2nd and 3rd third function calls result in a cache-hit. This is because every object passed in is identical.\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 <= inputs.length <= 10<sup>5</sup></code></li>\n\t<li><code>0 <= inputs.flat().length <= 10<sup>5</sup></code></li>\n\t<li><code>inputs[i][j] != NaN</code></li>\n</ul>\n",
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"translatedContent": "<p>请你编写一个函数，它接收一个函数参数 <code>fn</code>，并返回该函数的 <strong>记忆化</strong> 后的结果。</p>\n\n<p><strong>记忆函数</strong> 是一个对于相同的输入永远不会被调用两次的函数。相反，它将返回一个缓存值。</p>\n\n<p><code>fn</code> 可以是任何函数，对于它接受什么类型的值没有限制。如果输入为 <code>===</code>，则认为输入相同。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong> \ngetInputs = () => [[2,2],[2,2],[1,2]]\nfn = function (a, b) { return a + b; }\n<b>输出：</b>[{\"val\":4,\"calls\":1},{\"val\":4,\"calls\":1},{\"val\":3,\"calls\":2}]\n<strong>解释：</strong>\nconst inputs = getInputs();\nconst memoized = memoize(fn);\nfor (const arr of inputs) {\n memoized(...arr);\n}\n\n对于参数为 (2, 2) 的输入: 2 + 2 = 4，需要调用 fn() 。\n对于参数为 (2, 2) 的输入: 2 + 2 = 4，这些输入此前调用过，因此不需要调用 fn() 。\n对于参数为 (1, 2) 的输入: 1 + 2 = 3，需要再次调用 fn()，总调用数为 2 。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<b>输入：</b>\ngetInputs = () => [[{},{}],[{},{}],[{},{}]] \nfn = function (a, b) { return a + b; }\n<b>输出：</b>[{\"val\":{},\"calls\":1},{\"val\":{},\"calls\":2},{\"val\":{},\"calls\":3}]\n<strong>解释：</strong>\n合并两个空对象总是会得到一个空对象。因为缓存命中，所以只有 1 次对 fn() 的调用，尽管这些对象之间没有一个是相同的（===）。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong> \ngetInputs = () => { const o = {}; return [[o,o],[o,o],[o,o]]; }\nfn = function (a, b) { return ({...a, ...b}); }\n<b>输出：</b>[{\"val\":{},\"calls\":1},{\"val\":{},\"calls\":1},{\"val\":{},\"calls\":1}]\n<strong>解释：</strong>\n合并两个空对象总是会得到一个空对象。第 2 和第 3 个函数调用导致缓存命中。这是因为传入的每个对象都是相同的。\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 <= inputs.length <= 10<sup>5</sup></code></li>\n\t<li><code>0 <= inputs.flat().length <= 10<sup>5</sup></code></li>\n\t<li><code>inputs[i][j] != NaN</code></li>\n</ul>\n",
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63
leetcode-cn/originData/memoize.json
Normal file
63
leetcode-cn/originData/memoize.json
Normal file
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||||
"content": "<p>Given a function <code>fn</code>, return a <strong>memoized</strong> version of that function.</p>\n\n<p>A <strong>memoized </strong>function is a function that will never be called twice with the same inputs. Instead it will returned a cached value.</p>\n\n<p>You can assume there are <strong>3 </strong>possible input functions: <code>sum</code><strong>, </strong><code>fib</code><strong>, </strong>and <code>factorial</code><strong>.</strong></p>\n\n<ul>\n\t<li><code>sum</code><strong> </strong>accepts two integers <code>a</code> and <code>b</code> and returns <code>a + b</code>.</li>\n\t<li><code>fib</code><strong> </strong>accepts a single integer <code>n</code> and returns <code>1</code> if <font face=\"monospace\"><code>n <= 1</code> </font>or<font face=\"monospace\"> <code>fib(n - 1) + fib(n - 2)</code> </font>otherwise.</li>\n\t<li><code>factorial</code> accepts a single integer <code>n</code> and returns <code>1</code> if <code>n <= 1</code> or <code>factorial(n - 1) * n</code> otherwise.</li>\n</ul>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input</strong>\n"sum"\n["call","call","getCallCount","call","getCallCount"]\n[[2,2],[2,2],[],[1,2],[]]\n<strong>Output</strong>\n[4,4,1,3,2]\n\n<strong>Explanation</strong>\nconst sum = (a, b) => a + b;\nconst memoizedSum = memoize(sum);\nmemoizedSum(2, 2); // Returns 4. sum() was called as (2, 2) was not seen before.\nmemoizedSum(2, 2); // Returns 4. However sum() was not called because the same inputs were seen before.\n// Total call count: 1\nmemoizedSum(1, 2); // Returns 3. sum() was called as (1, 2) was not seen before.\n// Total call count: 2\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input\n</strong>"factorial"\n["call","call","call","getCallCount","call","getCallCount"]\n[[2],[3],[2],[],[3],[]]\n<strong>Output</strong>\n[2,6,2,2,6,2]\n\n<strong>Explanation</strong>\nconst factorial = (n) => (n <= 1) ? 1 : (n * factorial(n - 1));\nconst memoFactorial = memoize(factorial);\nmemoFactorial(2); // Returns 2.\nmemoFactorial(3); // Returns 6.\nmemoFactorial(2); // Returns 2. However factorial was not called because 2 was seen before.\n// Total call count: 2\nmemoFactorial(3); // Returns 6. However factorial was not called because 3 was seen before.\n// Total call count: 2\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input\n</strong>"fib"\n["call","getCallCount"]\n[[5],[]]\n<strong>Output</strong>\n[8,1]\n\n<strong>Explanation\n</strong>fib(5) = 8\n// Total call count: 1\n\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>0 <= a, b <= 10<sup>5</sup></code></li>\n\t<li><code>1 <= n <= 10</code></li>\n\t<li><code>at most 10<sup>5</sup> function calls</code></li>\n\t<li><code>at most 10<sup>5</sup> attempts to access callCount</code></li>\n\t<li><code>input function is sum, fib, or factorial</code></li>\n</ul>\n",
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||||
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"translatedContent": "<p>请你编写一个函数，它接收另一个函数作为输入，并返回该函数的 <strong>记忆化</strong> 后的结果。</p>\n\n<p><strong>记忆函数</strong> 是一个对于相同的输入永远不会被调用两次的函数。相反，它将返回一个缓存值。</p>\n\n<p>你可以假设有 <strong>3</strong> 个可能的输入函数：<code>sum</code> 、<code>fib</code> 和 <code>factorial</code> 。</p>\n\n<ul>\n\t<li> <code>sum</code> 接收两个整型参数 <code>a</code> 和 <code>b</code> ，并返回 <code>a + b</code> 。</li>\n\t<li> <code>fib</code> 接收一个整型参数 <code>n</code> ，如果 <code>n <= 1</code> 则返回 <code>1</code>，否则返回 <code>fib (n - 1) + fib (n - 2)</code>。</li>\n\t<li> <code>factorial</code> 接收一个整型参数 <code>n</code> ，如果 <code>n <= 1</code> 则返回 <code>1</code> ，否则返回 <code>factorial(n - 1) * n</code> 。</li>\n</ul>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>\n\"sum\"\n[\"call\",\"call\",\"getCallCount\",\"call\",\"getCallCount\"]\n[[2,2],[2,2],[],[1,2],[]]\n<strong>输出：</strong>\n[4,4,1,3,2]\n\n<strong>解释：</strong>\nconst sum = (a, b) => a + b;\nconst memoizedSum = memoize(sum);\nmemoizedSum (2, 2);// 返回 4。sum() 被调用，因为之前没有使用参数 (2, 2) 调用过。\nmemoizedSum (2, 2);// 返回 4。没有调用 sum()，因为前面有相同的输入。\n//总调用数： 1\nmemoizedSum(1、2);// 返回 3。sum() 被调用，因为之前没有使用参数 (1, 2) 调用过。\n//总调用数： 2\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：\n</strong>\"factorial\"\n[\"call\",\"call\",\"call\",\"getCallCount\",\"call\",\"getCallCount\"]\n[[2],[3],[2],[],[3],[]]\n<strong>输出：</strong>\n[2,6,2,2,6,2]\n\n<strong>解释：</strong>\nconst factorial = (n) => (n <= 1) ? 1 : (n * factorial(n - 1));\nconst memoFactorial = memoize(factorial);\nmemoFactorial(2); // 返回 2。\nmemoFactorial(3); // 返回 6。\nmemoFactorial(2); // 返回 2。 没有调用 factorial()，因为前面有相同的输入。\n// 总调用数：2\nmemoFactorial(3); // 返回 6。 没有调用 factorial()，因为前面有相同的输入。\n// 总调用数：2\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：\n</strong>\"fib\"\n[\"call\",\"getCallCount\"]\n[[5],[]]\n<strong>输出：</strong>\n[8,1]\n\n<strong>解释：\n</strong>fib(5) = 8\n// 总调用数：1\n\n</pre>\n\n<p> </p>\n\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>0 <= a, b <= 10<sup>5</sup></code></li>\n\t<li><code>1 <= n <= 10</code></li>\n\t<li><code>at most 10<sup>5</sup> function calls</code></li>\n\t<li><code>at most 10<sup>5</sup> attempts to access callCount</code></li>\n\t<li><code>input function is sum, fib, or factorial</code></li>\n</ul>\n",
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|
||||
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63
leetcode-cn/originData/nested-array-generator.json
Normal file
63
leetcode-cn/originData/nested-array-generator.json
Normal file
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"content": "<p>Given a <strong>multi-dimensional array</strong> of integers, return a generator object which yields integers in the same order as <strong>inorder traversal</strong>.</p>\n\n<p>A <strong>multi-dimensional array</strong> is a recursive data structure that contains both integers and other <strong>multi-dimensional arrays</strong>.</p>\n\n<p><strong>inorder traversal</strong> iterates over each array from left to right, yielding any integers it encounters or applying <strong>inorder traversal</strong> to any arrays it encounters.</p>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> arr = [[[6]],[1,3],[]]\n<strong>Output:</strong> [6,1,3]\n<strong>Explanation:</strong>\nconst generator = inorderTraversal(arr);\ngenerator.next().value; // 6\ngenerator.next().value; // 1\ngenerator.next().value; // 3\ngenerator.next().done; // true\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> arr = []\n<strong>Output:</strong> []\n<strong>Explanation:</strong> There are no integers so the generator doesn't yield anything.\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>0 <= arr.flat().length <= 10<sup>5</sup></code></li>\n\t<li><code>0 <= arr.flat()[i] <= 10<sup>5</sup></code></li>\n\t<li><code>maxNestingDepth <= 10<sup>5</sup></code></li>\n</ul>\n",
|
||||
"translatedTitle": "嵌套数组生成器",
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||||
"translatedContent": "<p>现给定一个整数的 <strong>多维数组</strong> ，请你返回一个生成器对象，按照 <strong>中序遍历</strong> 的顺序逐个生成整数。</p>\n\n<p><strong>多维数组</strong> 是一个递归数据结构，包含整数和其他 <strong>多维数组</strong>。</p>\n\n<p><strong>中序遍历</strong> 是从左到右遍历每个数组，在遇到任何整数时生成它，遇到任何数组时递归应用 <strong>中序遍历</strong> 。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<b>输入：</b>arr = [[[6]],[1,3],[]]\n<b>输出：</b>[6,1,3]\n<strong>解释：</strong>\nconst generator = inorderTraversal(arr);\ngenerator.next().value; // 6\ngenerator.next().value; // 1\ngenerator.next().value; // 3\ngenerator.next().done; // true\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<b>输入：</b>arr = []\n<b>输出：</b>[]\n<b>解释：</b>输入的多维数组没有任何参数，所以生成器不需要生成任何值。\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>0 <= arr.flat().length <= 10<sup>5</sup></code></li>\n\t<li><code>0 <= arr.flat()[i] <= 10<sup>5</sup></code></li>\n\t<li><code>maxNestingDepth <= 10<sup>5</sup></code></li>\n</ul>\n",
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||||
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||||
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62
leetcode-cn/originData/promise-pool.json
Normal file
62
leetcode-cn/originData/promise-pool.json
Normal file
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leetcode-cn/originData/promise-time-limit.json
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"content": "<p>Given an asyncronous function <code>fn</code> and a time <code>t</code> in milliseconds, return a new <strong>time limited</strong> version of the input function.</p>\n\n<p>A <strong>time limited</strong> function is a function that is identical to the original unless it takes longer than <code>t</code> milliseconds to fullfill. In that case, it will reject with <code>"Time Limit Exceeded"</code>. Note that it should reject with a string, not an <code>Error</code>.</p>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> \nfn = async (n) => { \n await new Promise(res => setTimeout(res, 100)); \n return n * n; \n}\ninputs = [5]\nt = 50\n<strong>Output:</strong> {"rejected":"Time Limit Exceeded","time":50}\n<strong>Explanation:</strong>\nThe provided function is set to resolve after 100ms. However, the time limit is set to 50ms. It rejects at t=50ms because the time limit was reached.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> \nfn = async (n) => { \n await new Promise(res => setTimeout(res, 100)); \n return n * n; \n}\ninputs = [5]\nt = 150\n<strong>Output:</strong> {"resolved":25,"time":100}\n<strong>Explanation:</strong>\nThe function resolved 5 * 5 = 25 at t=100ms. The time limit is never reached.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> \nfn = async (a, b) => { \n await new Promise(res => setTimeout(res, 120)); \n return a + b; \n}\ninputs = [5,10]\nt = 150\n<strong>Output:</strong> {"resolved":15,"time":120}\n<strong>Explanation:</strong>\nThe function resolved 5 + 10 = 15 at t=120ms. The time limit is never reached.\n</pre>\n\n<p><strong class=\"example\">Example 4:</strong></p>\n\n<pre>\n<strong>Input:</strong> \nfn = async () => { \n throw "Error";\n}\ninputs = []\nt = 1000\n<strong>Output:</strong> {"rejected":"Error","time":0}\n<strong>Explanation:</strong>\nThe function immediately throws an error.</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>0 <= inputs.length <= 10</code></li>\n\t<li><code>0 <= t <= 1000</code></li>\n\t<li><code>fn returns a promise</code></li>\n</ul>\n",
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"translatedContent": "<p>请你编写一个函数，它接收一个异步函数 <code>fn</code> 和一个以毫秒为单位的时间 <code>t</code>。它应根据限时函数返回一个有 <strong>限时</strong> 效果的函数。</p>\n\n<p>限时函数是与原函数相同的函数，除非它需要 <code>t</code> 毫秒以上的时间来完成。如果出现了这种情况，请你返回 <code>\"Time Limit Exceeded\"</code> 拒绝这次函数的调用。注意，它应该返回一个字符串拒绝，而不是一个 <code>Error</code> 。</p>\n\n<p> </p>\n\n<p><b>示例 1：</b></p>\n\n<pre>\n<b>输入：</b>\nfn = async (n) => { \n await new Promise(res => setTimeout(res, 100)); \n return n * n; \n}\ninputs = [5]\nt = 50\n<b>输出：</b>{\"rejected\":\"Time Limit Exceeded\",\"time\":50}\n<b>解释：\n</b>提供的函数设置在 100ms 后执行完成，但是设置的超时时间为 50ms，所以在 t=50ms 时拒绝因为达到了超时时间。\n</pre>\n\n<p><b>示例 2：</b></p>\n\n<pre>\n<b>输入：</b>\nfn = async (n) => { \n await new Promise(res => setTimeout(res, 100)); \n return n * n; \n}\ninputs = [5]\nt = 150\n<b>输出：</b>{\"resolved\":25,\"time\":100}\n<b>解释：</b>\n在 t=100ms 时执行 5*5=25 ，没有达到超时时间。\n</pre>\n\n<p><b>示例 3：</b></p>\n\n<pre>\n<b>输入：</b>\nfn = async (a, b) => { \n await new Promise(res => setTimeout(res, 120)); \n return a + b; \n}\ninputs = [5,10]\nt = 150\n<b>输出：</b>{\"resolved\":15,\"time\":120}\n<b>解释：\n</b>在 t=120ms 时执行 5+10=15，没有达到超时时间。\n</pre>\n\n<p><b>示例 4：</b></p>\n\n<pre>\n<b>输入：</b>\nfn = async () => { \n throw \"Error\";\n}\ninputs = []\nt = 1000\n<b>输出：</b>{\"rejected\":\"Error\",\"time\":0}\n<b>解释：</b>\n此函数始终丢出 Error</pre>\n\n<p> </p>\n\n<p><b>提示：</b></p>\n\n<ul>\n\t<li><code>0 <= inputs.length <= 10</code></li>\n\t<li><code>0 <= t <= 1000</code></li>\n\t<li><code>fn 返回一个 Promise 对象</code></li>\n</ul>\n",
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161
leetcode-cn/originData/rdmXM7.json
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161
leetcode-cn/originData/rdmXM7.json
Normal file
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176
leetcode-cn/originData/row-with-maximum-ones.json
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176
leetcode-cn/originData/row-with-maximum-ones.json
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File diff suppressed because one or more lines are too long
63
leetcode-cn/originData/sleep.json
Normal file
63
leetcode-cn/originData/sleep.json
Normal file
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"content": "<p>Given a positive integer <code>millis</code>, write an asyncronous function that sleeps for <code>millis</code> milliseconds. It can resolve any value.</p>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> millis = 100\n<strong>Output:</strong> 100\n<strong>Explanation:</strong> It should return a promise that resolves after 100ms.\nlet t = Date.now();\nsleep(100).then(() => {\n console.log(Date.now() - t); // 100\n});\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> millis = 200\n<strong>Output:</strong> 200\n<strong>Explanation:</strong> It should return a promise that resolves after 200ms.\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 <= millis <= 1000</code></li>\n</ul>\n",
|
||||
"translatedTitle": "睡眠函数",
|
||||
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||||
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||||
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||||
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164
leetcode-cn/originData/sliding-subarray-beauty.json
Normal file
164
leetcode-cn/originData/sliding-subarray-beauty.json
Normal file
File diff suppressed because one or more lines are too long
62
leetcode-cn/originData/snail-traversal.json
Normal file
62
leetcode-cn/originData/snail-traversal.json
Normal file
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||||
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||||
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||||
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||||
"content": "<p>Write code that enhances all arrays such that you can call the <code>snail(rowsCount, colsCount)</code> method that transforms the 1D array into a 2D array organised in the pattern known as <strong>snail traversal order</strong>. Invalid input values should output an empty array. If <code>rowsCount * colsCount !== nums.length</code>, the input is considered invalid.</p>\n\n<p><strong>Snail traversal order</strong><em> </em>starts at the top left cell with the first value of the current array. It then moves through the entire first column from top to bottom, followed by moving to the next column on the right and traversing it from bottom to top. This pattern continues, alternating the direction of traversal with each column, until the entire current array is covered. For example, when given the input array <code>[19, 10, 3, 7, 9, 8, 5, 2, 1, 17, 16, 14, 12, 18, 6, 13, 11, 20, 4, 15]</code> with <code>rowsCount = 5</code> and <code>colsCount = 4</code>, the desired output matrix is shown below. Note that iterating the matrix following the arrows corresponds to the order of numbers in the original array.</p>\n\n<p> </p>\n\n<p><img alt=\"Traversal Diagram\" src=\"https://assets.leetcode.com/uploads/2023/04/10/screen-shot-2023-04-10-at-100006-pm.png\" style=\"width: 275px; height: 343px;\" /></p>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> \nnums = [19, 10, 3, 7, 9, 8, 5, 2, 1, 17, 16, 14, 12, 18, 6, 13, 11, 20, 4, 15]\nrowsCount = 5\ncolsCount = 4\n<strong>Output:</strong> \n[\n [19,17,16,15],\n [10,1,14,4],\n [3,2,12,20],\n [7,5,18,11],\n [9,8,6,13]\n]\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> \nnums = [1,2,3,4]\nrowsCount = 1\ncolsCount = 4\n<strong>Output:</strong> [[1, 2, 3, 4]]\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> \nnums = [1,3]\nrowsCount = 2\ncolsCount = 2\n<strong>Output:</strong> []\n<strong>Explanation:</strong> 2 multiplied by 2 is 4, and the original array [1,3] has a length of 2; therefore, the input is invalid.\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>0 <= nums.length <= 250</code></li>\n\t<li><code>1 <= nums[i] <= 1000</code></li>\n\t<li><code>1 <= rowsCount <= 250</code></li>\n\t<li><code>1 <= colsCount <= 250</code></li>\n</ul>\n\n<p> </p>\n",
|
||||
"translatedTitle": "蜗牛排序",
|
||||
"translatedContent": "<p>请你编写一段代码为所有数组实现 <code>snail(rowsCount，colsCount)</code> 方法，该方法将 1D 数组转换为以蜗牛排序的模式的 2D 数组。无效的输入值应该输出一个空数组。当 <code>rowsCount * colsCount !==</code><code>nums.length</code> 时。这个输入被认为是无效的。</p>\n\n<p>蜗牛排序从左上角的单元格开始，从当前数组的第一个值开始。然后，它从上到下遍历第一列，接着移动到右边的下一列，并从下到上遍历它。将这种模式持续下去，每列交替变换遍历方向，直到覆盖整个数组。例如，当给定输入数组 <code>[19, 10, 3, 7, 9, 8, 5, 2, 1, 17, 16, 14, 12, 18, 6, 13, 11, 20, 4, 15]</code> ，当 <code>rowsCount = 5</code> 且 <code>colsCount = 4</code> 时，需要输出矩阵如下图所示。注意，矩阵沿箭头方向对应于原数组中数字的顺序</p>\n\n<p> </p>\n\n<p><img alt=\"Traversal Diagram\" src=\"https://assets.leetcode.com/uploads/2023/04/10/screen-shot-2023-04-10-at-100006-pm.png\" style=\"width: 275px; height: 343px;\" /></p>\n\n<p> </p>\n\n<p><b>示例 1：</b></p>\n\n<pre>\n<b>输入：</b>\nnums = [19, 10, 3, 7, 9, 8, 5, 2, 1, 17, 16, 14, 12, 18, 6, 13, 11, 20, 4, 15]\nrowsCount = 5\ncolsCount = 4\n<b>输出：</b>\n[\n [19,17,16,15],\n [10,1,14,4],\n [3,2,12,20],\n [7,5,18,11],\n [9,8,6,13]\n]\n</pre>\n\n<p><b>示例 2：</b></p>\n\n<pre>\n<b>输入：</b>\nnums = [1,2,3,4]\nrowsCount = 1\ncolsCount = 4\n<b>输出：</b>[[1, 2, 3, 4]]\n</pre>\n\n<p><b>示例 3：</b></p>\n\n<pre>\n<b>输入：</b>\nnums = [1,3]\nrowsCount = 2\ncolsCount = 2\n<b>输出：</b>[]\n<strong>Explanation:</strong> 2 * 2 = 4, 且原数组 [1,3] 的长度为 2; 所以，输入是无效的。\n</pre>\n\n<p> </p>\n\n<p><b>提示：</b></p>\n\n<ul>\n\t<li><code>0 <= nums.length <= 250</code></li>\n\t<li><code>1 <= nums[i] <= 1000</code></li>\n\t<li><code>1 <= rowsCount <= 250</code></li>\n\t<li><code>1 <= colsCount <= 250</code></li>\n</ul>\n",
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||||
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||||
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|
||||
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|
||||
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||||
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||||
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||||
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|
||||
"envInfo": "{\"javascript\":[\"JavaScript\",\"<p>\\u7248\\u672c\\uff1a<code>Node.js 16.13.2<\\/code><\\/p>\\r\\n\\r\\n<p>\\u60a8\\u7684\\u4ee3\\u7801\\u5728\\u6267\\u884c\\u65f6\\u5c06\\u5e26\\u4e0a <code>--harmony<\\/code> \\u6807\\u8bb0\\u6765\\u5f00\\u542f <a href=\\\"http:\\/\\/node.green\\/\\\" target=\\\"_blank\\\">\\u65b0\\u7248ES6\\u7279\\u6027<\\/a>\\u3002<\\/p>\\r\\n\\r\\n<p><a href=\\\"https:\\/\\/lodash.com\\\" target=\\\"_blank\\\">lodash.js<\\/a> \\u5e93\\u5df2\\u7ecf\\u9ed8\\u8ba4\\u88ab\\u5305\\u542b\\u3002<\\/p>\\r\\n\\r\\n<p> \\u5982\\u9700\\u4f7f\\u7528\\u961f\\u5217\\/\\u4f18\\u5148\\u961f\\u5217\\uff0c\\u60a8\\u53ef\\u4f7f\\u7528 <a href=\\\"https:\\/\\/github.com\\/datastructures-js\\/priority-queue\\/tree\\/fb4fdb984834421279aeb081df7af624d17c2a03\\\" target=\\\"_blank\\\"> datastructures-js\\/priority-queue@5.3.0<\\/a> \\u548c <a href=\\\"https:\\/\\/github.com\\/datastructures-js\\/queue\\/tree\\/e63563025a5a805aa16928cb53bcd517bfea9230\\\" target=\\\"_blank\\\"> datastructures-js\\/queue@4.2.1<\\/a>\\u3002<\\/p>\"],\"typescript\":[\"TypeScript\",\"<p>TypeScript 4.5.4<\\/p>\\r\\n\\r\\n<p>Compile Options: --alwaysStrict --strictBindCallApply --strictFunctionTypes --target ES2020<\\/p>\\r\\n\\r\\n<p><a href=\\\"https:\\/\\/lodash.com\\\" target=\\\"_blank\\\">lodash.js<\\/a> \\u5e93\\u5df2\\u7ecf\\u9ed8\\u8ba4\\u88ab\\u5305\\u542b\\u3002<\\/p>\\r\\n\\r\\n<p> \\u5982\\u9700\\u4f7f\\u7528\\u961f\\u5217\\/\\u4f18\\u5148\\u961f\\u5217\\uff0c\\u60a8\\u53ef\\u4f7f\\u7528 <a href=\\\"https:\\/\\/github.com\\/datastructures-js\\/priority-queue\\/tree\\/fb4fdb984834421279aeb081df7af624d17c2a03\\\" target=\\\"_blank\\\"> datastructures-js\\/priority-queue@5.3.0<\\/a> \\u548c <a href=\\\"https:\\/\\/github.com\\/datastructures-js\\/queue\\/tree\\/e63563025a5a805aa16928cb53bcd517bfea9230\\\" target=\\\"_blank\\\"> datastructures-js\\/queue@4.2.1<\\/a>\\u3002<\\/p>\"]}",
|
||||
"book": null,
|
||||
"isSubscribed": false,
|
||||
"isDailyQuestion": false,
|
||||
"dailyRecordStatus": null,
|
||||
"editorType": "CKEDITOR",
|
||||
"ugcQuestionId": null,
|
||||
"style": "LEETCODE",
|
||||
"exampleTestcases": "[19, 10, 3, 7, 9, 8, 5, 2, 1, 17, 16, 14, 12, 18, 6, 13, 11, 20, 4, 15]\n5\n4\n[1,2,3,4]\n1\n4\n[1,3]\n2\n2",
|
||||
"__typename": "QuestionNode"
|
||||
}
|
||||
}
|
||||
}
|
||||
163
leetcode-cn/originData/sum-multiples.json
Normal file
163
leetcode-cn/originData/sum-multiples.json
Normal file
File diff suppressed because one or more lines are too long
161
leetcode-cn/originData/xepqZ5.json
Normal file
161
leetcode-cn/originData/xepqZ5.json
Normal file
File diff suppressed because one or more lines are too long
@ -0,0 +1,73 @@
|
||||
<p>请你编写一个异步函数 <code>promisePool</code> ，它接收一个异步函数数组 <code>functions</code> 和 <strong>池限制</strong> <code>n</code>。它应该返回一个 promise 对象，当所有输入函数都执行完毕后，promise 对象就执行完毕。</p>
|
||||
|
||||
<p><strong>池限制</strong> 定义是一次可以挂起的最多 promise 对象的数量。<code>promisePool</code> 应该开始执行尽可能多的函数，并在旧的 promise 执行完毕后继续执行新函数。<code>promisePool</code> 应该先执行 <code>functions[i]</code>，再执行 <code>functions[i + 1]</code>，然后执行 <code>functions[i + 2]</code>，等等。当最后一个 promise 执行完毕时，<code>promisePool</code> 也应该执行完毕。</p>
|
||||
|
||||
<p>例如，如果 <code>n = 1</code> , <code>promisePool</code> 在序列中每次执行一个函数。然而，如果 <code>n = 2</code> ，它首先执行两个函数。当两个函数中的任何一个执行完毕后，再执行第三个函数(如果它是可用的)，依此类推，直到没有函数要执行为止。</p>
|
||||
|
||||
<p>你可以假设所有的 <code>functions</code> 都不会被拒绝。对于 <code>promisePool</code> 来说，返回一个可以解析任何值的 promise 都是可以接受的。</p>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong class="example">示例 1：</strong></p>
|
||||
|
||||
<pre>
|
||||
<b>输入：</b>
|
||||
functions = [
|
||||
() => new Promise(res => setTimeout(res, 300)),
|
||||
() => new Promise(res => setTimeout(res, 400)),
|
||||
() => new Promise(res => setTimeout(res, 200))
|
||||
]
|
||||
n = 2
|
||||
<b>输出：</b>[[300,400,500],500]
|
||||
<strong>解释</strong>
|
||||
传递了三个函数。它们的睡眠时间分别为 300ms、 400ms 和 200ms。
|
||||
在 t=0 时，执行前两个函数。池大小限制达到 2。
|
||||
当 t=300 时，第一个函数执行完毕后，执行第3个函数。池大小为 2。
|
||||
在 t=400 时，第二个函数执行完毕后。没有什么可执行的了。池大小为 1。
|
||||
在 t=500 时，第三个函数执行完毕后。池大小为 0，因此返回的 promise 也执行完成。
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">示例 2：</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>输入：
|
||||
</strong>functions = [
|
||||
() => new Promise(res => setTimeout(res, 300)),
|
||||
() => new Promise(res => setTimeout(res, 400)),
|
||||
() => new Promise(res => setTimeout(res, 200))
|
||||
]
|
||||
n = 5
|
||||
<b>输出：</b>[[300,400,200],400]
|
||||
<strong>解释：</strong>
|
||||
在 t=0 时，所有3个函数都被执行。池的限制大小 5 永远不会满足。
|
||||
在 t=200 时，第三个函数执行完毕后。池大小为 2。
|
||||
在 t=300 时，第一个函数执行完毕后。池大小为 1。
|
||||
在 t=400 时，第二个函数执行完毕后。池大小为 0，因此返回的 promise 也执行完成。
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">示例 3：</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>输入：</strong>
|
||||
functions = [
|
||||
() => new Promise(res => setTimeout(res, 300)),
|
||||
() => new Promise(res => setTimeout(res, 400)),
|
||||
() => new Promise(res => setTimeout(res, 200))
|
||||
]
|
||||
n = 1
|
||||
<b>输出：</b>[[300,700,900],900]
|
||||
<strong>解释：</strong>
|
||||
在 t=0 时，执行第一个函数。池大小为1。
|
||||
当 t=300 时，第一个函数执行完毕后，执行第二个函数。池大小为 1。
|
||||
当 t=700 时，第二个函数执行完毕后，执行第三个函数。池大小为 1。
|
||||
在 t=900 时，第三个函数执行完毕后。池大小为 0，因此返回的 Promise 也执行完成。
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>提示：</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>0 <= functions.length <= 10</code></li>
|
||||
<li><code><font face="monospace">1 <= n <= 10</font></code></li>
|
||||
</ul>
|
||||
@ -0,0 +1,45 @@
|
||||
<p>定义一个数组 <code>arr</code> 的 <strong>转换数组</strong> <code>conver</code> 为：</p>
|
||||
|
||||
<ul>
|
||||
<li><code>conver[i] = arr[i] + max(arr[0..i])</code>，其中 <code>max(arr[0..i])</code> 是满足 <code>0 <= j <= i</code> 的所有 <code>arr[j]</code> 中的最大值。</li>
|
||||
</ul>
|
||||
|
||||
<p>定义一个数组 <code>arr</code> 的 <strong>分数</strong> 为 <code>arr</code> 转换数组中所有元素的和。</p>
|
||||
|
||||
<p>给你一个下标从 <strong>0</strong> 开始长度为 <code>n</code> 的整数数组 <code>nums</code> ，请你返回一个长度为 <code>n</code> 的数组<em> </em><code>ans</code> ，其中 <code>ans[i]</code>是前缀 <code>nums[0..i]</code> 的分数。</p>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>示例 1：</strong></p>
|
||||
|
||||
<pre><b>输入：</b>nums = [2,3,7,5,10]
|
||||
<b>输出：</b>[4,10,24,36,56]
|
||||
<b>解释：</b>
|
||||
对于前缀 [2] ，转换数组为 [4] ，所以分数为 4 。
|
||||
对于前缀 [2, 3] ，转换数组为 [4, 6] ，所以分数为 10 。
|
||||
对于前缀 [2, 3, 7] ，转换数组为 [4, 6, 14] ，所以分数为 24 。
|
||||
对于前缀 [2, 3, 7, 5] ，转换数组为 [4, 6, 14, 12] ，所以分数为 36 。
|
||||
对于前缀 [2, 3, 7, 5, 10] ，转换数组为 [4, 6, 14, 12, 20] ，所以分数为 56 。
|
||||
</pre>
|
||||
|
||||
<p><strong>示例 2：</strong></p>
|
||||
|
||||
<pre><b>输入：</b>nums = [1,1,2,4,8,16]
|
||||
<b>输出：</b>[2,4,8,16,32,64]
|
||||
<b>解释：</b>
|
||||
对于前缀 [1] ，转换数组为 [2] ，所以分数为 2 。
|
||||
对于前缀 [1, 1]，转换数组为 [2, 2] ，所以分数为 4 。
|
||||
对于前缀 [1, 1, 2]，转换数组为 [2, 2, 4] ，所以分数为 8 。
|
||||
对于前缀 [1, 1, 2, 4]，转换数组为 [2, 2, 4, 8] ，所以分数为 16 。
|
||||
对于前缀 [1, 1, 2, 4, 8]，转换数组为 [2, 2, 4, 8, 16] ，所以分数为 32 。
|
||||
对于前缀 [1, 1, 2, 4, 8, 16]，转换数组为 [2, 2, 4, 8, 16, 32] ，所以分数为 64 。
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>提示：</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>1 <= nums.length <= 10<sup>5</sup></code></li>
|
||||
<li><code>1 <= nums[i] <= 10<sup>9</sup></code></li>
|
||||
</ul>
|
||||
@ -0,0 +1,41 @@
|
||||
<p>给你一个大小为 <code>m x n</code> 的二进制矩阵 <code>mat</code> ，请你找出包含最多 <strong>1</strong> 的行的下标（从 <strong>0</strong> 开始）以及这一行中 <strong>1</strong> 的数目。</p>
|
||||
|
||||
<p>如果有多行包含最多的 1 ，只需要选择 <strong>行下标最小</strong> 的那一行。</p>
|
||||
|
||||
<p>返回一个由行下标和该行中 1 的数量组成的数组。</p>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>示例 1：</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>输入：</strong>mat = [[0,1],[1,0]]
|
||||
<strong>输出：</strong>[0,1]
|
||||
<strong>解释：</strong>两行中 1 的数量相同。所以返回下标最小的行，下标为 0 。该行 1 的数量为 1 。所以，答案为 [0,1] 。
|
||||
</pre>
|
||||
|
||||
<p><strong>示例 2：</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>输入：</strong>mat = [[0,0,0],[0,1,1]]
|
||||
<strong>输出：</strong>[1,2]
|
||||
<strong>解释：</strong>下标为 1 的行中 1 的数量最多<code>。</code>该行 1 的数量<code>为 2 。所以，答案为</code> [1,2] 。
|
||||
</pre>
|
||||
|
||||
<p><strong>示例 3：</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>输入：</strong>mat = [[0,0],[1,1],[0,0]]
|
||||
<strong>输出：</strong>[1,2]
|
||||
<strong>解释：</strong>下标为 1 的行中 1 的数量最多。该行 1 的数量<code>为 2 。所以，答案为</code> [1,2] 。</pre>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>提示：</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>m == mat.length</code> </li>
|
||||
<li><code>n == mat[i].length</code> </li>
|
||||
<li><code>1 <= m, n <= 100</code> </li>
|
||||
<li><code>mat[i][j]</code> 为 <code>0</code> 或 <code>1</code></li>
|
||||
</ul>
|
||||
@ -0,0 +1,47 @@
|
||||
<p>给你一棵二叉树的根 <code>root</code> ，请你将每个节点的值替换成该节点的所有 <strong>堂兄弟节点值的和 </strong>。</p>
|
||||
|
||||
<p>如果两个节点在树中有相同的深度且它们的父节点不同，那么它们互为 <strong>堂兄弟</strong> 。</p>
|
||||
|
||||
<p>请你返回修改值之后，树的根<em> </em><code>root</code><em> </em>。</p>
|
||||
|
||||
<p><strong>注意</strong>，一个节点的深度指的是从树根节点到这个节点经过的边数。</p>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>示例 1：</strong></p>
|
||||
|
||||
<p><img alt="" src="https://assets.leetcode.com/uploads/2023/01/11/example11.png" style="width: 571px; height: 151px;" /></p>
|
||||
|
||||
<pre>
|
||||
<b>输入：</b>root = [5,4,9,1,10,null,7]
|
||||
<b>输出：</b>[0,0,0,7,7,null,11]
|
||||
<b>解释：</b>上图展示了初始的二叉树和修改每个节点的值之后的二叉树。
|
||||
- 值为 5 的节点没有堂兄弟，所以值修改为 0 。
|
||||
- 值为 4 的节点没有堂兄弟，所以值修改为 0 。
|
||||
- 值为 9 的节点没有堂兄弟，所以值修改为 0 。
|
||||
- 值为 1 的节点有一个堂兄弟，值为 7 ，所以值修改为 7 。
|
||||
- 值为 10 的节点有一个堂兄弟，值为 7 ，所以值修改为 7 。
|
||||
- 值为 7 的节点有两个堂兄弟，值分别为 1 和 10 ，所以值修改为 11 。
|
||||
</pre>
|
||||
|
||||
<p><strong>示例 2：</strong></p>
|
||||
|
||||
<p><img alt="" src="https://assets.leetcode.com/uploads/2023/01/11/diagram33.png" style="width: 481px; height: 91px;" /></p>
|
||||
|
||||
<pre>
|
||||
<b>输入：</b>root = [3,1,2]
|
||||
<b>输出：</b>[0,0,0]
|
||||
<b>解释：</b>上图展示了初始的二叉树和修改每个节点的值之后的二叉树。
|
||||
- 值为 3 的节点没有堂兄弟，所以值修改为 0 。
|
||||
- 值为 1 的节点没有堂兄弟，所以值修改为 0 。
|
||||
- 值为 2 的节点没有堂兄弟，所以值修改为 0 。
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>提示：</strong></p>
|
||||
|
||||
<ul>
|
||||
<li>树中节点数目的范围是 <code>[1, 10<sup>5</sup>]</code> 。</li>
|
||||
<li><code>1 <= Node.val <= 10<sup>4</sup></code></li>
|
||||
</ul>
|
||||
60
leetcode-cn/problem (Chinese)/传送卷轴 [rdmXM7].md
Normal file
60
leetcode-cn/problem (Chinese)/传送卷轴 [rdmXM7].md
Normal file
@ -0,0 +1,60 @@
|
||||
随着不断的深入，小扣来到了守护者之森寻找的魔法水晶。首先，他必须先通过守护者的考验。
|
||||
|
||||
考验的区域是一个正方形的迷宫，`maze[i][j]` 表示在迷宫 `i` 行 `j` 列的地形：
|
||||
- 若为 `.` ，表示可以到达的空地；
|
||||
- 若为 `#` ，表示不可到达的墙壁；
|
||||
- 若为 `S` ，表示小扣的初始位置；
|
||||
- 若为 `T` ，表示魔法水晶的位置。
|
||||
|
||||
小扣每次可以向 上、下、左、右 相邻的位置移动一格。而守护者拥有一份「传送魔法卷轴」，使用规则如下：
|
||||
- 魔法需要在小扣位于 **空地** 时才能释放，发动后卷轴消失；；
|
||||
- 发动后，小扣会被传送到水平或者竖直的镜像位置，且目标位置不得为墙壁(如下图所示)；
|
||||
{:width=400px}
|
||||
|
||||
在使用卷轴后，小扣将被「附加负面效果」，因此小扣需要尽可能缩短传送后到达魔法水晶的距离。而守护者的目标是阻止小扣到达魔法水晶的位置；如果无法阻止，则尽可能 **增加** 小扣传送后到达魔法水晶的距离。
|
||||
假设小扣和守护者都按最优策略行事，返回小扣需要在 「附加负面效果」的情况下 **最少** 移动多少次才能到达魔法水晶。如果无法到达，返回 `-1`。
|
||||
|
||||
**注意：**
|
||||
- 守护者可以不使用卷轴；
|
||||
- 传送后的镜像位置可能与原位置相同。
|
||||
|
||||
**示例 1：**
|
||||
>输入：`maze = [".....","##S..","...#.","T.#..","###.."]`
|
||||
>
|
||||
>输出：`7`
|
||||
>
|
||||
>解释：如下图所示：
|
||||
>守护者释放魔法的两个最佳的位置为 [2,0] 或 [3,1]：
|
||||
>若小扣经过 [2,0]，守护者在该位置释放魔法，
|
||||
>小扣被传送至 [2,4] 处且加上负面效果，此时小扣还需要移动 7 次才能到达魔法水晶；
|
||||
>若小扣经过 [3,1]，守护者在该位置释放魔法，
|
||||
>小扣被传送至 [3,3] 处且加上负面效果，此时小扣还需要移动 9 次才能到达魔法水晶；
|
||||
>因此小扣负面效果下最少需要移动 7 次才能到达魔法水晶。
|
||||
{:width=300px}
|
||||
|
||||
|
||||
**示例 2：**
|
||||
>输入：`maze = [".#..","..##",".#S.",".#.T"]`
|
||||
>
|
||||
>输出：`-1`
|
||||
>
|
||||
>解释：如下图所示。
|
||||
>若小扣向下移动至 [3,2]，守护者使其传送至 [0,2]，小扣将无法到达魔法水晶；
|
||||
>若小扣向右移动至 [2,3]，守护者使其传送至 [2,0]，小扣将无法到达魔法水晶；
|
||||
{:width=300px}
|
||||
|
||||
|
||||
**示例 3：**
|
||||
>输入：`maze = ["S###.","..###","#..##","##..#","###.T"]`
|
||||
>
|
||||
>输出：`5`
|
||||
>
|
||||
>解释：如下图所示：
|
||||
>守护者需要小扣在空地才能释放，因此初始无法将其从 [0,0] 传送至 [0,4];
|
||||
>当小扣移动至 [2,1] 时，释放卷轴将其传送至水平方向的镜像位置 [2,1]（为原位置）
|
||||
>而后小扣需要移动 5 次到达魔法水晶
|
||||
{:width=300px}
|
||||
|
||||
**提示：**
|
||||
- `4 <= maze.length == maze[i].length <= 200`
|
||||
- `maze[i][j]` 仅包含 `"."`、`"#"`、`"S"`、`"T"`
|
||||
@ -0,0 +1,38 @@
|
||||
<p>给你一个下标从 <strong>0</strong> 开始的 <strong>正</strong> 整数数组 <code>nums</code> 。你可以对数组执行以下操作 <strong>任意</strong> 次：</p>
|
||||
|
||||
<ul>
|
||||
<li>选择一个满足 <code>0 <= i < n - 1</code> 的下标 <code>i</code> ，将 <code>nums[i]</code> 或者 <code>nums[i+1]</code> 两者之一替换成它们的最大公约数。</li>
|
||||
</ul>
|
||||
|
||||
<p>请你返回使数组 <code>nums</code> 中所有元素都等于 <code>1</code> 的 <strong>最少</strong> 操作次数。如果无法让数组全部变成 <code>1</code> ，请你返回 <code>-1</code> 。</p>
|
||||
|
||||
<p>两个正整数的最大公约数指的是能整除这两个数的最大正整数。</p>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>示例 1：</strong></p>
|
||||
|
||||
<pre><b>输入：</b>nums = [2,6,3,4]
|
||||
<b>输出：</b>4
|
||||
<b>解释：</b>我们可以执行以下操作：
|
||||
- 选择下标 i = 2 ，将 nums[2] 替换为 gcd(3,4) = 1 ，得到 nums = [2,6,1,4] 。
|
||||
- 选择下标 i = 1 ，将 nums[1] 替换为 gcd(6,1) = 1 ，得到 nums = [2,1,1,4] 。
|
||||
- 选择下标 i = 0 ，将 nums[0] 替换为 gcd(2,1) = 1 ，得到 nums = [1,1,1,4] 。
|
||||
- 选择下标 i = 2 ，将 nums[3] 替换为 gcd(1,4) = 1 ，得到 nums = [1,1,1,1] 。
|
||||
</pre>
|
||||
|
||||
<p><strong>示例 2：</strong></p>
|
||||
|
||||
<pre><b>输入：</b>nums = [2,10,6,14]
|
||||
<b>输出：</b>-1
|
||||
<b>解释：</b>无法将所有元素都变成 1 。
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>提示：</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>2 <= nums.length <= 50</code></li>
|
||||
<li><code>1 <= nums[i] <= 10<sup>6</sup></code></li>
|
||||
</ul>
|
||||
34
leetcode-cn/problem (Chinese)/倍数求和 [sum-multiples].html
Normal file
34
leetcode-cn/problem (Chinese)/倍数求和 [sum-multiples].html
Normal file
@ -0,0 +1,34 @@
|
||||
<p>给你一个正整数 <code>n</code> ，请你计算在 <code>[1，n]</code> 范围内能被 <code>3</code>、<code>5</code>、<code>7</code> 整除的所有整数之和。</p>
|
||||
|
||||
<p>返回一个整数，用于表示给定范围内所有满足约束条件的数字之和。</p>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>示例 1：</strong></p>
|
||||
|
||||
<pre><strong>输入：</strong>n = 7
|
||||
<strong>输出：</strong>21
|
||||
<strong>解释：</strong>在 <code>[1, 7]</code> 范围内能被 3、<code>5、</code><code>7 整除的所有整数分别是</code><code> 3、5、6、7</code> 。数字之和为 <code>21</code> 。
|
||||
</pre>
|
||||
|
||||
<p><strong>示例 2：</strong></p>
|
||||
|
||||
<pre><strong>输入：</strong>n = 10
|
||||
<strong>输出：</strong>40
|
||||
<strong>解释：</strong>在 <code>[1, 10]</code> 范围内能被 3、<code>5、</code><code>7 整除的所有整数分别是</code><code> 3、5、6、7、9、10</code> 。数字之和为 <code>40</code> 。
|
||||
</pre>
|
||||
|
||||
<p><strong>示例 3：</strong></p>
|
||||
|
||||
<pre><strong>输入：</strong>n = 9
|
||||
<strong>输出：</strong>30
|
||||
<strong>解释：</strong>在 <code>[1, 9]</code> 范围内能被 3、<code>5、</code><code>7 整除的所有整数分别是</code><code> 3、5、6、7、9</code> 。数字之和为 <code>30</code> 。
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>提示：</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>1 <= n <= 10<sup>3</sup></code></li>
|
||||
</ul>
|
||||
79
leetcode-cn/problem (Chinese)/函数防抖 [debounce].html
Normal file
79
leetcode-cn/problem (Chinese)/函数防抖 [debounce].html
Normal file
@ -0,0 +1,79 @@
|
||||
<p>请你编写一个函数，接收参数为另一个函数和一个以毫秒为单位的时间 <code>t</code> ，并返回该函数的 <b>函数防抖 </b>后的结果。</p>
|
||||
|
||||
<p><b>函数防抖 </b>方法是一个函数，它的执行被延迟了 <code>t</code> 毫秒，如果在这个时间窗口内再次调用它，它的执行将被取消。你编写的防抖函数也应该接收传递的参数。</p>
|
||||
|
||||
<p>例如，假设 <code>t = 50ms</code> ，函数分别在 <code>30ms</code> 、 <code>60ms</code> 和 <code>100ms</code> 时调用。前两个函数调用将被取消，第三个函数调用将在 <code>150ms</code> 执行。如果改为 <code>t = 35ms</code> ，则第一个调用将被取消，第二个调用将在 <code>95ms</code> 执行，第三个调用将在 <code>135ms</code> 执行。</p>
|
||||
|
||||
<p><img alt="Debounce Schematic" src="https://assets.leetcode.com/uploads/2023/04/08/screen-shot-2023-04-08-at-11048-pm.png" style="width: 800px; height: 242px;" /></p>
|
||||
|
||||
<p>上图展示了了防抖函数是如何转换事件的。其中，每个矩形表示 100ms，反弹时间为 400ms。每种颜色代表一组不同的输入。</p>
|
||||
|
||||
<p>请在不使用 lodash 的 <code>_.debounce()</code> 函数的前提下解决该问题。</p>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong class="example">示例 1：</strong></p>
|
||||
|
||||
<pre>
|
||||
<b>输入：</b>
|
||||
t = 50
|
||||
calls = [
|
||||
{"t": 50, inputs: [1]},
|
||||
{"t": 75, inputs: [2]}
|
||||
]
|
||||
<b>输出：</b>[{"t": 125, inputs: [2]}]
|
||||
<strong>解释：</strong>
|
||||
let start = Date.now();
|
||||
function log(...inputs) {
|
||||
console.log([Date.now() - start, inputs ])
|
||||
}
|
||||
const dlog = debounce(log, 50);
|
||||
setTimeout(() => dlog(1), 50);
|
||||
setTimeout(() => dlog(2), 75);
|
||||
|
||||
第一次调用被第二次调用取消，因为第二次调用发生在 100ms 之前
|
||||
第二次调用延迟 50ms，在 125ms 执行。输入为 (2)。
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">示例 2：</strong></p>
|
||||
|
||||
<pre>
|
||||
<b>输入：</b>
|
||||
t = 20
|
||||
calls = [
|
||||
{"t": 50, inputs: [1]},
|
||||
{"t": 100, inputs: [2]}
|
||||
]
|
||||
<b>输出：</b>[{"t": 70, inputs: [1]}, {"t": 120, inputs: [2]}]
|
||||
<strong>解释：</strong>
|
||||
第一次调用延迟到 70ms。输入为 (1)。
|
||||
第二次调用延迟到 120ms。输入为 (2)。
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">示例 3：</strong></p>
|
||||
|
||||
<pre>
|
||||
<b>输入：</b>
|
||||
t = 150
|
||||
calls = [
|
||||
{"t": 50, inputs: [1, 2]},
|
||||
{"t": 300, inputs: [3, 4]},
|
||||
{"t": 300, inputs: [5, 6]}
|
||||
]
|
||||
<b>输出：</b>[{"t": 200, inputs: [1,2]}, {"t": 450, inputs: [5, 6]}]
|
||||
<strong>解释：</strong>
|
||||
第一次调用延迟了 150ms，运行时间为 200ms。输入为 (1, 2)。
|
||||
第二次调用被第三次调用取消
|
||||
第三次调用延迟了 150ms，运行时间为 450ms。输入为 (5, 6)。
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>提示：</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>0 <= t <= 1000</code></li>
|
||||
<li><code>1 <= calls.length <= 10</code></li>
|
||||
<li><code>0 <= calls[i].t <= 1000</code></li>
|
||||
<li><code>0 <= calls[i].inputs.length <= 10</code></li>
|
||||
</ul>
|
||||
85
leetcode-cn/problem (Chinese)/分组 [group-by].html
Normal file
85
leetcode-cn/problem (Chinese)/分组 [group-by].html
Normal file
@ -0,0 +1,85 @@
|
||||
<p>请你编写一段可应用于所有数组的代码，使任何数组调用 <code>array. groupBy(fn)</code> 方法时，它返回对该数组 <strong>分组后</strong> 的结果。</p>
|
||||
|
||||
<p>数组 <strong>分组</strong> 是一个对象，其中的每个键都是 <code>fn(arr[i])</code> 的输出的一个数组，该数组中含有原数组中具有该键的所有项。</p>
|
||||
|
||||
<p>提供的回调函数 <code>fn</code> 将接受数组中的项并返回一个字符串类型的键。</p>
|
||||
|
||||
<p>每个值列表的顺序应该与元素在数组中出现的顺序相同。任何顺序的键都是可以接受的。</p>
|
||||
|
||||
<p>请在不使用 lodash 的 <code>_.groupBy</code> 函数的前提下解决这个问题。</p>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>示例 1：</strong></p>
|
||||
|
||||
<pre>
|
||||
<b>输入：</b>
|
||||
array = [
|
||||
{"id":"1"},
|
||||
{"id":"1"},
|
||||
{"id":"2"}
|
||||
],
|
||||
fn = function (item) {
|
||||
return item.id;
|
||||
}
|
||||
<b>输出：</b>
|
||||
{
|
||||
"1": [{"id": "1"}, {"id": "1"}],
|
||||
"2": [{"id": "2"}]
|
||||
}
|
||||
<strong>解释：</strong>
|
||||
输出来自函数 array.groupBy(fn)。
|
||||
分组选择方法是从数组中的每个项中获取 "id" 。
|
||||
有两个 "id" 为 1 的对象。所以将这两个对象都放在第一个数组中。
|
||||
有一个 "id" 为 2 的对象。所以该对象被放到第二个数组中。
|
||||
</pre>
|
||||
|
||||
<p><strong>示例 2：</strong></p>
|
||||
|
||||
<pre>
|
||||
<b>输入：</b>
|
||||
array = [
|
||||
[1, 2, 3],
|
||||
[1, 3, 5],
|
||||
[1, 5, 9]
|
||||
]
|
||||
fn = function (list) {
|
||||
return String(list[0]);
|
||||
}
|
||||
<b>输出：</b>
|
||||
{
|
||||
"1": [[1, 2, 3], [1, 3, 5], [1, 5, 9]]
|
||||
}
|
||||
<strong>解释：</strong>
|
||||
数组可以是任何类型的。在本例中，分组选择方法是将键定义为数组中的第一个元素。
|
||||
所有数组的第一个元素都是1，所以它们被组合在一起。
|
||||
{
|
||||
"1": [[1, 2, 3], [1, 3, 5], [1, 5, 9]]
|
||||
}
|
||||
</pre>
|
||||
|
||||
<p><strong>示例 3：</strong></p>
|
||||
|
||||
<pre>
|
||||
<b>输出：</b>
|
||||
array = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
|
||||
fn = function (n) {
|
||||
return String(n > 5);
|
||||
}
|
||||
<strong>输入：</strong>
|
||||
{
|
||||
"true": [6, 7, 8, 9, 10],
|
||||
"false": [1, 2, 3, 4, 5]
|
||||
}
|
||||
<strong>解释：</strong>
|
||||
分组选择方法是根据每个数字是否大于 5 来分割数组。
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>提示：</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>0 <= array.length <= 10<sup>5</sup></code></li>
|
||||
<li><code>fn 返回一个字符串</code></li>
|
||||
</ul>
|
||||
@ -0,0 +1,52 @@
|
||||
<p>请你编写一个函数，它接收一个函数数组 <code>[f1, f2, f3，…]， fn]</code> ，并返回一个新的函数 <code>fn</code> ，它是函数数组的 <strong>复合函数</strong> 。</p>
|
||||
|
||||
<p><code>[f(x)， g(x)， h(x)]</code> 的 <strong>复合函数</strong> 为 <code>fn(x) = f(g(h(x)))</code> 。</p>
|
||||
|
||||
<p>一个空函数列表的 <strong>复合函数</strong> 是 <strong>恒等函数</strong> <code>f(x) = x</code> 。</p>
|
||||
|
||||
<p>你可以假设数组中的每个函数接受一个整型参数作为输入，并返回一个整型作为输出。</p>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong class="example">示例 1：</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>输入：</strong>functions = [x => x + 1, x => x * x, x => 2 * x], x = 4
|
||||
<b>输出：</b>65
|
||||
<strong>解释：</strong>
|
||||
从右向左计算......
|
||||
Starting with x = 4.
|
||||
2 * (4) = 8
|
||||
(8) * (8) = 64
|
||||
(64) + 1 = 65
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">示例 2：</strong></p>
|
||||
|
||||
<pre>
|
||||
<b>输出：</b>functions = [x => 10 * x, x => 10 * x, x => 10 * x], x = 1
|
||||
<b>输入：</b>1000
|
||||
<strong>解释：</strong>
|
||||
从右向左计算......
|
||||
10 * (1) = 10
|
||||
10 * (10) = 100
|
||||
10 * (100) = 1000
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">示例 3：</strong></p>
|
||||
|
||||
<pre>
|
||||
<b>输入：</b>functions = [], x = 42
|
||||
<b>输出：</b>42
|
||||
<strong>解释：</strong>
|
||||
空函数列表的复合函数就是恒等函数</pre>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>提示：</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code><font face="monospace">-1000 <= x <= 1000</font></code></li>
|
||||
<li><code><font face="monospace">0 <= functions.length <= 1000</font></code></li>
|
||||
<li><font face="monospace"><code>所有函数都接受并返回一个整型</code></font></li>
|
||||
</ul>
|
||||
@ -0,0 +1,50 @@
|
||||
<p>给定两个对象 <code>o1</code> 和 <code>o2</code> ，请你检查它们是否 <strong>完全相等</strong> 。</p>
|
||||
|
||||
<p>对于两个 <strong>完全相等</strong> 的对象，它们必须包含相同的键，并且相关的值也必须 <strong>完全相等</strong> 。如果两个对象通过了 <code>===</code> 相等性检查，它们也被认为是 <strong>完全相等</strong> 的。</p>
|
||||
|
||||
<p>你可以假设这两个对象都是 <code>JSON.parse</code> 的输出。换句话说，它们是有效的 <code>JSON</code> 。</p>
|
||||
|
||||
<p>请你在不使用 lodash 的 <code>_.isEqual()</code> 函数的前提下解决这个问题。</p>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>示例 1：</strong></p>
|
||||
|
||||
<pre>
|
||||
<b>输入：</b>o1 = {"x":1,"y":2}, o2 = {"x":1,"y":2}
|
||||
<b>输出：</b>true
|
||||
<b>输入：</b>键和值完全匹配。
|
||||
</pre>
|
||||
|
||||
<p><strong>示例 2：</strong></p>
|
||||
|
||||
<pre>
|
||||
<b>输入：</b>o1 = {"y":2,"x":1}, o2 = {"x":1,"y":2}
|
||||
<b>输出：</b>true
|
||||
<b>解释：</b>尽管键的顺序不同，但它们仍然完全匹配。
|
||||
</pre>
|
||||
|
||||
<p><strong>示例 3：</strong></p>
|
||||
|
||||
<pre>
|
||||
<b>输入：</b>o1 = {"x":null,"L":[1,2,3]}, o2 = {"x":null,"L":["1","2","3"]}
|
||||
<b>输出：</b>false
|
||||
<b>解释：</b>数字数组不同于字符串数组。
|
||||
</pre>
|
||||
|
||||
<p><strong>示例 4：</strong></p>
|
||||
|
||||
<pre>
|
||||
<b>输入：</b>o1 = true, o2 = false
|
||||
<b>输出：</b>false
|
||||
<b>解释：</b>true !== false</pre>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>提示：</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>1 <= JSON.stringify(o1).length <= 10<sup>5</sup></code></li>
|
||||
<li><code>1 <= JSON.stringify(o2).length <= 10<sup>5</sup></code></li>
|
||||
<li><code>maxNestingDepth <= 1000</code></li>
|
||||
</ul>
|
||||
@ -0,0 +1,50 @@
|
||||
<p>现给定一个对象，返回该对象的有效 JSON 字符串。你可以假设这个对象只包括字符串、整数、数组、对象、布尔值和 null。返回的字符串不能包含额外的空格。键的返回顺序应该与 <code>Object.keys()</code> 的顺序相同。</p>
|
||||
|
||||
<p>请你在不使用内置方法 <code>JSON.stringify</code> 的前提下解决这个问题。</p>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>示例 1：</strong></p>
|
||||
|
||||
<pre>
|
||||
<b>输入：</b>object = {"y":1,"x":2}
|
||||
<b>输出：</b>{"y":1,"x":2}
|
||||
<b>解释：</b>
|
||||
返回该对象的 JSON 表示形式。
|
||||
注意，键的返回顺序应该与 Object.keys() 的顺序相同。</pre>
|
||||
|
||||
<p><strong>示例 2：</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>输入：</strong>object = {"a":"str","b":-12,"c":true,"d":null}
|
||||
<b>输出：</b>{"a":"str","b":-12,"c":true,"d":null}
|
||||
<strong>解释：</strong>
|
||||
JSON 的基本类型是字符串、数字型、布尔值和 null。
|
||||
</pre>
|
||||
|
||||
<p><strong>示例 3：</strong></p>
|
||||
|
||||
<pre>
|
||||
<b>输入：</b>object = {"key":{"a":1,"b":[{},null,"Hello"]}}
|
||||
<b>输出：</b>{"key":{"a":1,"b":[{},null,"Hello"]}}
|
||||
<b>解释：</b>
|
||||
对象和数组可以包括其他对象和数组。
|
||||
</pre>
|
||||
|
||||
<p><strong>示例 4：</strong></p>
|
||||
|
||||
<pre>
|
||||
<b>输入：</b>object = true
|
||||
<b>输出：</b>true
|
||||
<b>解释：</b>
|
||||
基本类型是有效的输入</pre>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>Constraints:</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>对象包括字符串、整数、布尔值、数组、对象和 null</code></li>
|
||||
<li><code>1 <= JSON.stringify(object).length <= 10<sup>5</sup></code></li>
|
||||
<li><code>maxNestingLevel <= 1000</code></li>
|
||||
</ul>
|
||||
@ -0,0 +1,38 @@
|
||||
<p>现给定一个整数的 <strong>多维数组</strong> ，请你返回一个生成器对象，按照 <strong>中序遍历</strong> 的顺序逐个生成整数。</p>
|
||||
|
||||
<p><strong>多维数组</strong> 是一个递归数据结构，包含整数和其他 <strong>多维数组</strong>。</p>
|
||||
|
||||
<p><strong>中序遍历</strong> 是从左到右遍历每个数组，在遇到任何整数时生成它，遇到任何数组时递归应用 <strong>中序遍历</strong> 。</p>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>示例 1：</strong></p>
|
||||
|
||||
<pre>
|
||||
<b>输入：</b>arr = [[[6]],[1,3],[]]
|
||||
<b>输出：</b>[6,1,3]
|
||||
<strong>解释：</strong>
|
||||
const generator = inorderTraversal(arr);
|
||||
generator.next().value; // 6
|
||||
generator.next().value; // 1
|
||||
generator.next().value; // 3
|
||||
generator.next().done; // true
|
||||
</pre>
|
||||
|
||||
<p><strong>示例 2：</strong></p>
|
||||
|
||||
<pre>
|
||||
<b>输入：</b>arr = []
|
||||
<b>输出：</b>[]
|
||||
<b>解释：</b>输入的多维数组没有任何参数，所以生成器不需要生成任何值。
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>提示：</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>0 <= arr.flat().length <= 10<sup>5</sup></code></li>
|
||||
<li><code>0 <= arr.flat()[i] <= 10<sup>5</sup></code></li>
|
||||
<li><code>maxNestingDepth <= 10<sup>5</sup></code></li>
|
||||
</ul>
|
||||
@ -0,0 +1,51 @@
|
||||
<p>给你两个下标从 <strong>0</strong> 开始的整数数组 <code>nums</code> 和 <code>divisors</code> 。</p>
|
||||
|
||||
<p><code>divisors[i]</code> 的 <strong>可整除性得分</strong> 等于满足 <code>nums[j]</code> 能被 <code>divisors[i]</code> 整除的下标 <code>j</code> 的数量。</p>
|
||||
|
||||
<p>返回 <strong>可整除性得分</strong> 最大的整数 <code>divisors[i]</code> 。如果有多个整数具有最大得分，则返回数值最小的一个。</p>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>示例 1：</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>输入：</strong>nums = [4,7,9,3,9], divisors = [5,2,3]
|
||||
<strong>输出：</strong>3
|
||||
<strong>解释：</strong>divisors 中每个元素的可整除性得分为：
|
||||
divisors[0] 的可整除性得分为 0 ，因为 nums 中没有任何数字能被 5 整除。
|
||||
divisors[1] 的可整除性得分为 1 ，因为 nums[0] 能被 2 整除。
|
||||
divisors[2] 的可整除性得分为 3 ，因为 nums[2]、nums[3] 和 nums[4] 都能被 3 整除。
|
||||
因此，返回 divisors[2] ，它的可整除性得分最大。
|
||||
</pre>
|
||||
|
||||
<p><strong>示例 2：</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>输入：</strong>nums = [20,14,21,10], divisors = [5,7,5]
|
||||
<strong>输出：</strong>5
|
||||
<strong>解释：</strong>divisors 中每个元素的可整除性得分为：
|
||||
divisors[0] 的可整除性得分为 2 ，因为 nums[0] 和 nums[3] 都能被 5 整除。
|
||||
divisors[1] 的可整除性得分为 2 ，因为 nums[1] 和 nums[2] 都能被 7 整除。
|
||||
divisors[2] 的可整除性得分为 2 ，因为 nums[0] 和 nums[3] 都能被5整除。
|
||||
由于 divisors[0]、divisors[1] 和 divisors[2] 的可整除性得分都是最大的，因此，我们返回数值最小的一个，即 divisors[2] 。
|
||||
</pre>
|
||||
|
||||
<p><strong>示例 3：</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>输入：</strong>nums = [12], divisors = [10,16]
|
||||
<strong>输出：</strong>10
|
||||
<strong>解释：</strong>divisors 中每个元素的可整除性得分为：
|
||||
divisors[0] 的可整除性得分为 0 ，因为 nums 中没有任何数字能被 10 整除。
|
||||
divisors[1] 的可整除性得分为 0 ，因为 nums 中没有任何数字能被 16 整除。
|
||||
由于 divisors[0] 和 divisors[1] 的可整除性得分都是最大的，因此，我们返回数值最小的一个，即 divisors[0] 。
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>提示：</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>1 <= nums.length, divisors.length <= 1000</code></li>
|
||||
<li><code>1 <= nums[i], divisors[i] <= 10<sup>9</sup></code></li>
|
||||
</ul>
|
||||
50
leetcode-cn/problem (Chinese)/探险营地 [0Zeoeg].md
Normal file
50
leetcode-cn/problem (Chinese)/探险营地 [0Zeoeg].md
Normal file
@ -0,0 +1,50 @@
|
||||
探险家小扣的行动轨迹，都将保存在记录仪中。`expeditions[i]` 表示小扣第 `i` 次探险记录，用一个字符串数组表示。其中的每个「营地」由大小写字母组成，通过子串 `->` 连接。
|
||||
> 例："Leet->code->Campsite"，表示到访了 "Leet"、"code"、"Campsite" 三个营地。
|
||||
|
||||
`expeditions[0]` 包含了初始小扣已知的所有营地；对于之后的第 `i` 次探险(即 `expeditions[i]` 且 i > 0)，如果记录中包含了之前均没出现的营地，则表示小扣 **新发现** 的营地。
|
||||
|
||||
请你找出小扣发现新营地最多且索引最小的那次探险，并返回对应的记录索引。如果所有探险记录都没有发现新的营地，返回 `-1`
|
||||
|
||||
**注意：**
|
||||
- 大小写不同的营地视为不同的营地；
|
||||
- 营地的名称长度均大于 `0`。
|
||||
|
||||
**示例 1：**
|
||||
>输入：`expeditions = ["leet->code","leet->code->Campsite->Leet","leet->code->leet->courier"]`
|
||||
>
|
||||
>输出：`1`
|
||||
>
|
||||
>解释：
|
||||
>初始已知的所有营地为 "leet" 和 "code"
|
||||
>第 1 次，到访了 "leet"、"code"、"Campsite"、"Leet"，新发现营地 2 处："Campsite"、"Leet"
|
||||
>第 2 次，到访了 "leet"、"code"、"courier"，新发现营地 1 处："courier"
|
||||
>第 1 次探险发现的新营地数量最多，因此返回 `1`
|
||||
|
||||
**示例 2：**
|
||||
>输入：`expeditions = ["Alice->Dex","","Dex"]`
|
||||
>
|
||||
>输出：`-1`
|
||||
>
|
||||
>解释：
|
||||
>初始已知的所有营地为 "Alice" 和 "Dex"
|
||||
>第 1 次，未到访任何营地；
|
||||
>第 2 次，到访了 "Dex"，未新发现营地；
|
||||
>因为两次探险均未发现新的营地，返回 `-1`
|
||||
|
||||
**示例 3：**
|
||||
>输入：`expeditions = ["","Gryffindor->Slytherin->Gryffindor","Hogwarts->Hufflepuff->Ravenclaw"]`
|
||||
>
|
||||
>输出：`2`
|
||||
>
|
||||
>解释：
|
||||
>初始未发现任何营地；
|
||||
>第 1 次，到访 "Gryffindor"、"Slytherin" 营地，其中重复到访 "Gryffindor" 两次，
|
||||
>因此新发现营地为 2 处："Gryffindor"、"Slytherin"
|
||||
>第 2 次，到访 "Hogwarts"、"Hufflepuff"、"Ravenclaw" 营地；
|
||||
>新发现营地 3 处："Hogwarts"、"Hufflepuff"、"Ravenclaw"；
|
||||
>第 2 次探险发现的新营地数量最多，因此返回 `2`
|
||||
|
||||
**提示：**
|
||||
- `1 <= expeditions.length <= 1000`
|
||||
- `0 <= expeditions[i].length <= 1000`
|
||||
- 探险记录中只包含大小写字母和子串"->"
|
||||
@ -0,0 +1,28 @@
|
||||
<p>请你编写一段代码实现一个数组方法，使任何数组都可以调用 <code>array.last()</code> 方法，这个方法将返回数组最后一个元素。如果数组中没有元素，则返回 <code>-1</code> 。</p>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>示例 1 ：</strong></p>
|
||||
|
||||
<pre>
|
||||
<b>输入：</b>nums = [1,2,3]
|
||||
<b>输出：</b>3
|
||||
<b>解释</b>：调用 nums.last() 后返回最后一个元素： 3。
|
||||
</pre>
|
||||
|
||||
<p><strong>示例 2 ：</strong></p>
|
||||
|
||||
<pre>
|
||||
<b>输入：</b>nums = []
|
||||
<b>输出：</b>-1
|
||||
<strong>解释：</strong>因为此数组没有元素，所以应该返回 -1。
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><b>提示：</b></p>
|
||||
|
||||
<ul>
|
||||
<li><code>0 <= arr.length <= 1000</code></li>
|
||||
<li><code>0 <= arr[i] <= 1000</code></li>
|
||||
</ul>
|
||||
@ -0,0 +1,64 @@
|
||||
<p>请你编写一个函数，它的参数为一个整数数组 <code>nums</code> 、一个计算函数 <code>fn</code> 和初始值 <font color="#c7254e"><font face="Menlo, Monaco, Consolas, Courier New, monospace"><span style="font-size:12.6px"><span style="background-color:#f9f2f4">init </span></span></font></font>。返回一个数组 <strong>归约后 </strong>的值。</p>
|
||||
|
||||
<p>你可以定义一个数组 <strong>归约后 </strong>的值，然后应用以下操作： <code>val = fn(init, nums[0])</code> ， <code>val = fn(val, nums[1])</code> ， <code>val = fn(val, arr[2])</code> ，<code>...</code> 直到数组中的每个元素都被处理完毕。返回 <code>val</code> 的最终值。</p>
|
||||
|
||||
<p>如果数组的长度为 0，它应该返回 <code>init</code> 的值。</p>
|
||||
|
||||
<p>请你在不使用内置数组方法的 <code>Array.reduce</code> 前提下解决这个问题。</p>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong class="example">示例 1：</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>输入：</strong>
|
||||
nums = [1,2,3,4]
|
||||
fn = function sum(accum, curr) { return accum + curr; }
|
||||
init = 0
|
||||
<strong>输出：</strong>10
|
||||
<strong>解释：</strong>
|
||||
初始值为 init=0 。
|
||||
(0) + nums[0] = 1
|
||||
(1) + nums[1] = 3
|
||||
(3) + nums[2] = 6
|
||||
(6) + nums[3] = 10
|
||||
Val 最终值为 10。
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">示例 2：</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>输入：</strong>
|
||||
nums = [1,2,3,4]
|
||||
fn = function sum(accum, curr) { return accum + curr * curr; }
|
||||
init = 100
|
||||
<strong>输出：</strong>130
|
||||
<strong>解释：</strong>
|
||||
初始值为 init=0 。
|
||||
(100) + nums[0]^2 = 101
|
||||
(101) + nums[1]^2 = 105
|
||||
(105) + nums[2]^2 = 114
|
||||
(114) + nums[3]^2 = 130
|
||||
Val 最终值为 130。
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">示例3:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>输入：</strong>
|
||||
nums = []
|
||||
fn = function sum(accum, curr) { return 0; }
|
||||
init = 25
|
||||
<strong>输出：</strong>25
|
||||
<b>解释：</b>这是一个空数组，所以返回 init 。
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>提示：</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>0 <= nums.length <= 1000</code></li>
|
||||
<li><code>0 <= nums[i] <= 1000</code></li>
|
||||
<li><code>0 <= init <= 1000</code></li>
|
||||
</ul>
|
||||
@ -0,0 +1,50 @@
|
||||
<p>现有一棵无向、无根的树，树中有 <code>n</code> 个节点，按从 <code>0</code> 到 <code>n - 1</code> 编号。给你一个整数 <code>n</code> 和一个长度为 <code>n - 1</code> 的二维整数数组 <code>edges</code> ，其中 <code>edges[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> 表示树中节点 <code>a<sub>i</sub></code> 和 <code>b<sub>i</sub></code> 之间存在一条边。</p>
|
||||
|
||||
<p>每个节点都关联一个价格。给你一个整数数组 <code>price</code> ，其中 <code>price[i]</code> 是第 <code>i</code> 个节点的价格。</p>
|
||||
|
||||
<p>给定路径的 <strong>价格总和</strong> 是该路径上所有节点的价格之和。</p>
|
||||
|
||||
<p>另给你一个二维整数数组 <code>trips</code> ，其中 <code>trips[i] = [start<sub>i</sub>, end<sub>i</sub>]</code> 表示您从节点 <code>start<sub>i</sub></code> 开始第 <code>i</code> 次旅行，并通过任何你喜欢的路径前往节点 <code>end<sub>i</sub></code> 。</p>
|
||||
|
||||
<p>在执行第一次旅行之前，你可以选择一些 <strong>非相邻节点</strong> 并将价格减半。</p>
|
||||
|
||||
<p>返回执行所有旅行的最小价格总和。</p>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>示例 1：</strong></p>
|
||||
<img alt="" src="https://assets.leetcode.com/uploads/2023/03/16/diagram2.png" style="width: 541px; height: 181px;">
|
||||
<pre><strong>输入：</strong>n = 4, edges = [[0,1],[1,2],[1,3]], price = [2,2,10,6], trips = [[0,3],[2,1],[2,3]]
|
||||
<strong>输出：</strong>23
|
||||
<strong>解释：
|
||||
</strong>上图表示将节点 2 视为根之后的树结构。第一个图表示初始树，第二个图表示选择节点 0 、2 和 3 并使其价格减半后的树。
|
||||
第 1 次旅行，选择路径 [0,1,3] 。路径的价格总和为 1 + 2 + 3 = 6 。
|
||||
第 2 次旅行，选择路径 [2,1] 。路径的价格总和为 2 + 5 = 7 。
|
||||
第 3 次旅行，选择路径 [2,1,3] 。路径的价格总和为 5 + 2 + 3 = 10 。
|
||||
所有旅行的价格总和为 6 + 7 + 10 = 23 。可以证明，23 是可以实现的最小答案。</pre>
|
||||
|
||||
<p><strong>示例 2：</strong></p>
|
||||
<img alt="" src="https://assets.leetcode.com/uploads/2023/03/16/diagram3.png" style="width: 456px; height: 111px;">
|
||||
<pre><strong>输入：</strong>n = 2, edges = [[0,1]], price = [2,2], trips = [[0,0]]
|
||||
<strong>输出：</strong>1
|
||||
<strong>解释：</strong>
|
||||
上图表示将节点 0 视为根之后的树结构。第一个图表示初始树，第二个图表示选择节点 0 并使其价格减半后的树。
|
||||
第 1 次旅行，选择路径 [0] 。路径的价格总和为 1 。
|
||||
所有旅行的价格总和为 1 。可以证明，1 是可以实现的最小答案。
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>提示：</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>1 <= n <= 50</code></li>
|
||||
<li><code>edges.length == n - 1</code></li>
|
||||
<li><code>0 <= a<sub>i</sub>, b<sub>i</sub> <= n - 1</code></li>
|
||||
<li><code>edges</code> 表示一棵有效的树</li>
|
||||
<li><code>price.length == n</code></li>
|
||||
<li><code>price[i]</code> 是一个偶数</li>
|
||||
<li><code>1 <= price[i] <= 1000</code></li>
|
||||
<li><code>1 <= trips.length <= 100</code></li>
|
||||
<li><code>0 <= start<sub>i</sub>, end<sub>i</sub> <= n - 1</code></li>
|
||||
</ul>
|
||||
35
leetcode-cn/problem (Chinese)/最强祝福力场 [xepqZ5].md
Normal file
35
leetcode-cn/problem (Chinese)/最强祝福力场 [xepqZ5].md
Normal file
@ -0,0 +1,35 @@
|
||||
小扣在探索丛林的过程中，无意间发现了传说中“落寞的黄金之都”。而在这片建筑废墟的地带中，小扣使用探测仪监测到了存在某种带有「祝福」效果的力场。
|
||||
经过不断的勘测记录，小扣将所有力场的分布都记录了下来。`forceField[i] = [x,y,side]` 表示第 `i` 片力场将覆盖以坐标 `(x,y)` 为中心，边长为 `side` 的正方形区域。
|
||||
|
||||
若任意一点的 **力场强度** 等于覆盖该点的力场数量，请求出在这片地带中 **力场强度** 最强处的 **力场强度**。
|
||||
|
||||
**注意：**
|
||||
- 力场范围的边缘同样被力场覆盖。
|
||||
|
||||
**示例 1：**
|
||||
>输入：
|
||||
>`forceField = [[0,0,1],[1,0,1]]`
|
||||
>
|
||||
>输出：`2`
|
||||
>
|
||||
>解释：如图所示，（0.5, 0) 处力场强度最强为 2， （0.5，-0.5）处力场强度同样是 2。
|
||||
{:width=400px}
|
||||
|
||||
|
||||
**示例 2：**
|
||||
>输入：
|
||||
>`forceField = [[4,4,6],[7,5,3],[1,6,2],[5,6,3]]`
|
||||
>
|
||||
>输出：`3`
|
||||
>
|
||||
>解释：如下图所示，
|
||||
>`forceField[0]、forceField[1]、forceField[3]` 重叠的区域力场强度最大，返回 `3`
|
||||
{:width=500px}
|
||||
|
||||
|
||||
|
||||
**提示：**
|
||||
- `1 <= forceField.length <= 100`
|
||||
- `forceField[i].length == 3`
|
||||
- `0 <= forceField[i][0], forceField[i][1] <= 10^9`
|
||||
- `1 <= forceField[i][2] <= 10^9`
|
||||
@ -0,0 +1,73 @@
|
||||
<p>请你编写一个函数，它接收一个异步函数 <code>fn</code> 和一个以毫秒为单位的时间 <code>t</code>。它应根据限时函数返回一个有 <strong>限时</strong> 效果的函数。</p>
|
||||
|
||||
<p>限时函数是与原函数相同的函数，除非它需要 <code>t</code> 毫秒以上的时间来完成。如果出现了这种情况，请你返回 <code>"Time Limit Exceeded"</code> 拒绝这次函数的调用。注意，它应该返回一个字符串拒绝，而不是一个 <code>Error</code> 。</p>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><b>示例 1：</b></p>
|
||||
|
||||
<pre>
|
||||
<b>输入：</b>
|
||||
fn = async (n) => {
|
||||
await new Promise(res => setTimeout(res, 100));
|
||||
return n * n;
|
||||
}
|
||||
inputs = [5]
|
||||
t = 50
|
||||
<b>输出：</b>{"rejected":"Time Limit Exceeded","time":50}
|
||||
<b>解释：
|
||||
</b>提供的函数设置在 100ms 后执行完成，但是设置的超时时间为 50ms，所以在 t=50ms 时拒绝因为达到了超时时间。
|
||||
</pre>
|
||||
|
||||
<p><b>示例 2：</b></p>
|
||||
|
||||
<pre>
|
||||
<b>输入：</b>
|
||||
fn = async (n) => {
|
||||
await new Promise(res => setTimeout(res, 100));
|
||||
return n * n;
|
||||
}
|
||||
inputs = [5]
|
||||
t = 150
|
||||
<b>输出：</b>{"resolved":25,"time":100}
|
||||
<b>解释：</b>
|
||||
在 t=100ms 时执行 5*5=25 ，没有达到超时时间。
|
||||
</pre>
|
||||
|
||||
<p><b>示例 3：</b></p>
|
||||
|
||||
<pre>
|
||||
<b>输入：</b>
|
||||
fn = async (a, b) => {
|
||||
await new Promise(res => setTimeout(res, 120));
|
||||
return a + b;
|
||||
}
|
||||
inputs = [5,10]
|
||||
t = 150
|
||||
<b>输出：</b>{"resolved":15,"time":120}
|
||||
<b>解释：
|
||||
</b>在 t=120ms 时执行 5+10=15，没有达到超时时间。
|
||||
</pre>
|
||||
|
||||
<p><b>示例 4：</b></p>
|
||||
|
||||
<pre>
|
||||
<b>输入：</b>
|
||||
fn = async () => {
|
||||
throw "Error";
|
||||
}
|
||||
inputs = []
|
||||
t = 1000
|
||||
<b>输出：</b>{"rejected":"Error","time":0}
|
||||
<b>解释：</b>
|
||||
此函数始终丢出 Error</pre>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><b>提示：</b></p>
|
||||
|
||||
<ul>
|
||||
<li><code>0 <= inputs.length <= 10</code></li>
|
||||
<li><code>0 <= t <= 1000</code></li>
|
||||
<li><code>fn 返回一个 Promise 对象</code></li>
|
||||
</ul>
|
||||
@ -0,0 +1,58 @@
|
||||
<p>编写一个类，它允许获取和设置键-值对，并且每个键都有一个 <strong>过期时间</strong> 。</p>
|
||||
|
||||
<p>该类有三个公共方法：</p>
|
||||
|
||||
<p><code>set(key, value, duration)</code> ：接收参数为整型键 <code>key</code> 、整型值 <code>value</code> 和以毫秒为单位的持续时间 <code>duration</code> 。一旦 <code>duration</code> 到期后，这个键就无法访问。如果相同的未过期键已经存在，该方法将返回 <code>true</code> ，否则返回 <code>false</code> 。如果该键已经存在，则它的值和持续时间都应该被覆盖。</p>
|
||||
|
||||
<p><code>get(key)</code> ：如果存在一个未过期的键，它应该返回这个键相关的值。否则返回 <code>-1</code> 。</p>
|
||||
|
||||
<p><code>count()</code> ：返回未过期键的总数。</p>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>示例 1：</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>输入：</strong>
|
||||
["TimeLimitedCache", "set", "get", "count", "get"]
|
||||
[[], [1, 42, 100], [1], [], [1]]
|
||||
[0, 0, 50, 50, 150]
|
||||
<strong>输出：</strong> [null, false, 42, 1, -1]
|
||||
<strong>解释：</strong>
|
||||
在 t=0 时，缓存被构造。
|
||||
在 t=0 时，添加一个键值对 (1: 42) ，过期时间为 100ms 。因为该值不存在，因此返回false。
|
||||
在 t=50 时，请求 key=1 并返回值 42。
|
||||
在 t=50 时，调用 count() ，缓存中有一个未过期的键。
|
||||
在 t=100 时，key=1 到期。
|
||||
在 t=150 时，调用 get(1) ，返回 -1，因为缓存是空的。
|
||||
</pre>
|
||||
|
||||
<p><strong>示例 2：</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>输入：</strong>
|
||||
["TimeLimitedCache", "set", "set", "get", "get", "get", "count"]
|
||||
[[], [1, 42, 50], [1, 50, 100], [1], [1], [1], []]
|
||||
[0, 0, 40, 50, 120, 200, 250]
|
||||
<strong>输出：</strong> [null, false, true, 50, 50, -1]
|
||||
<strong>解释：</strong>
|
||||
在 t=0 时，缓存被构造。
|
||||
在 t=0 时，添加一个键值对 (1: 42) ，过期时间为 50ms。因为该值不存在，因此返回false。
|
||||
当 t=40 时，添加一个键值对 (1: 50) ，过期时间为 100ms。因为一个未过期的键已经存在，返回 true 并覆盖这个键的旧值。
|
||||
在 t=50 时，调用 get(1) ，返回 50。
|
||||
在 t=120 时，调用 get(1) ，返回 50。
|
||||
在 t=140 时，key=1 过期。
|
||||
在 t=200 时，调用 get(1) ，但缓存为空，因此返回 -1。
|
||||
在 t=250 时，count() 返回0 ，因为缓存是空的，没有未过期的键。
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>提示：</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>0 <= key <= 10<sup>9</sup></code></li>
|
||||
<li><code>0 <= value <= 10<sup>9</sup></code></li>
|
||||
<li><code>0 <= duration <= 1000</code></li>
|
||||
<li><code>方法调用总数不会超过100</code></li>
|
||||
</ul>
|
||||
@ -0,0 +1,35 @@
|
||||
<p>给你一个字符串 <code>word</code> ，你可以向其中任何位置插入 "a"、"b" 或 "c" 任意次，返回使 <code>word</code> <strong>有效</strong> 需要插入的最少字母数。</p>
|
||||
|
||||
<p>如果字符串可以由 "abc" 串联多次得到，则认为该字符串 <strong>有效</strong> 。</p>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>示例 1：</strong></p>
|
||||
|
||||
<pre><strong>输入：</strong>word = "b"
|
||||
<strong>输出：</strong>2
|
||||
<strong>解释：</strong>在 "b" 之前插入 "a" ，在 "b" 之后插入 "c" 可以得到有效字符串 "<strong>a</strong>b<strong>c</strong>" 。
|
||||
</pre>
|
||||
|
||||
<p><strong>示例 2：</strong></p>
|
||||
|
||||
<pre><strong>输入：</strong>word = "aaa"
|
||||
<strong>输出：</strong>6
|
||||
<strong>解释：</strong>在每个 "a" 之后依次插入 "b" 和 "c" 可以得到有效字符串 "a<strong>bc</strong>a<strong>bc</strong>a<strong>bc</strong>" 。
|
||||
</pre>
|
||||
|
||||
<p><strong>示例 3：</strong></p>
|
||||
|
||||
<pre><strong>输入：</strong>word = "abc"
|
||||
<strong>输出：</strong>0
|
||||
<strong>解释：</strong>word 已经是有效字符串，不需要进行修改。
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>提示：</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>1 <= word.length <= 50</code></li>
|
||||
<li><code>word</code> 仅由字母 "a"、"b" 和 "c" 组成。</li>
|
||||
</ul>
|
||||
@ -0,0 +1,39 @@
|
||||
<p>给你一个下标从 <strong>0</strong> 开始的 <code>m x n</code> 整数矩阵 <code>grid</code> 。矩阵中某一列的宽度是这一列数字的最大 <strong>字符串长度</strong> 。</p>
|
||||
|
||||
<ul>
|
||||
<li>比方说，如果 <code>grid = [[-10], [3], [12]]</code> ，那么唯一一列的宽度是 <code>3</code> ，因为 <code>-10</code> 的字符串长度为 <code>3</code> 。</li>
|
||||
</ul>
|
||||
|
||||
<p>请你返回一个大小为 <code>n</code> 的整数数组 <code>ans</code> ，其中 <code>ans[i]</code> 是第 <code>i</code> 列的宽度。</p>
|
||||
|
||||
<p>一个有 <code>len</code> 个数位的整数 <code>x</code> ，如果是非负数，那么 <strong>字符串</strong><strong>长度</strong> 为 <code>len</code> ，否则为 <code>len + 1</code> 。</p>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>示例 1：</strong></p>
|
||||
|
||||
<pre><b>输入：</b>grid = [[1],[22],[333]]
|
||||
<b>输出：</b>[3]
|
||||
<b>解释：</b>第 0 列中，333 字符串长度为 3 。
|
||||
</pre>
|
||||
|
||||
<p><strong>示例 2：</strong></p>
|
||||
|
||||
<pre><b>输入：</b>grid = [[-15,1,3],[15,7,12],[5,6,-2]]
|
||||
<b>输出：</b>[3,1,2]
|
||||
<b>解释：</b>
|
||||
第 0 列中，只有 -15 字符串长度为 3 。
|
||||
第 1 列中，所有整数的字符串长度都是 1 。
|
||||
第 2 列中，12 和 -2 的字符串长度都为 2 。
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>提示：</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>m == grid.length</code></li>
|
||||
<li><code>n == grid[i].length</code></li>
|
||||
<li><code>1 <= m, n <= 100 </code></li>
|
||||
<li><code>-10<sup>9</sup> <= grid[r][c] <= 10<sup>9</sup></code></li>
|
||||
</ul>
|
||||
67
leetcode-cn/problem (Chinese)/柯里化 [curry].html
Normal file
67
leetcode-cn/problem (Chinese)/柯里化 [curry].html
Normal file
@ -0,0 +1,67 @@
|
||||
<p>请你编写一个函数，它接收一个其他的函数，并返回该函数的 <strong>柯里化 </strong>后的形式。</p>
|
||||
|
||||
<p><strong>柯里化 </strong>函数的定义是接受与原函数相同数量或更少数量的参数，并返回另一个 <strong>柯里化</strong> 后的函数或与原函数相同的值。</p>
|
||||
|
||||
<p>实际上，当你调用原函数，如 <code>sum(1,2,3)</code> 时，它将调用 <strong>柯里化</strong> 函数的某个形式，如 <code>csum(1)(2)(3)</code>， <code>csum(1)(2,3)</code>， <code>csum(1,2)(3)</code>，或 <code>csum(1,2,3)</code> 。所有调用 <strong>柯里化</strong> 函数的方法都应该返回与原始函数相同的值。</p>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>示例 1：</strong></p>
|
||||
|
||||
<pre>
|
||||
<b>输入：</b>
|
||||
fn = function sum(a, b, c) { return a + b + c; }
|
||||
inputs = [[1],[2],[3]]
|
||||
<b>输出：</b>6
|
||||
<strong>解释：</strong>
|
||||
执行的代码是：
|
||||
const curriedSum = curry(fn);
|
||||
curriedSum(1)(2)(3) === 6;
|
||||
curriedSum(1)(2)(3) 应该返回像原函数 sum(1, 2, 3) 一样的值。
|
||||
</pre>
|
||||
|
||||
<p><strong>示例 2：</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>输入：</strong>
|
||||
fn = function sum(a, b, c) { return a + b + c; }
|
||||
inputs = [[1,2],[3]]]
|
||||
<b>输出：</b>6
|
||||
<strong>解释：</strong>
|
||||
curriedSum(1, 2)(3) 应该返回像原函数 sum(1, 2, 3) 一样的值。</pre>
|
||||
|
||||
<p><strong>示例 3：</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>输入：</strong>
|
||||
fn = function sum(a, b, c) { return a + b + c; }
|
||||
inputs = [[],[],[1,2,3]]
|
||||
<b>输出：</b>6
|
||||
<strong>解释：</strong>
|
||||
你应该能够以任何方式传递参数，包括一次性传递所有参数或完全不传递参数。
|
||||
curriedSum()()(1, 2, 3) 应该返回像原函数 sum(1, 2, 3) 一样的值。
|
||||
</pre>
|
||||
|
||||
<p><strong>示例 4：</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>输入：</strong>
|
||||
fn = function life() { return 42; }
|
||||
inputs = [[]]
|
||||
<b>输出：</b>42
|
||||
<strong>解释：</strong>
|
||||
柯里化一个没有接收参数，没做有效操作的函数。
|
||||
curriedLife() === 42
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>提示：</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>1 <= inputs.length <= 1000</code></li>
|
||||
<li><code>0 <= inputs[i][j] <= 10<sup>5</sup></code></li>
|
||||
<li><code>0 <= fn.length <= 1000</code></li>
|
||||
<li><code>inputs.flat().length == fn.length</code></li>
|
||||
<li><code>函数参数需要被显式定义</code></li>
|
||||
</ul>
|
||||
@ -0,0 +1,40 @@
|
||||
<p>请你编写一个函数，检查给定的对象是否是给定类或超类的实例。</p>
|
||||
|
||||
<p>可以传递给函数的数据类型没有限制。</p>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>示例 1：</strong></p>
|
||||
|
||||
<pre>
|
||||
<b>输入：</b>func = () => checkIfInstance(new Date(), Date)
|
||||
<b>输出：</b>true
|
||||
<strong>解释：</strong>根据定义，Date 构造函数返回的对象是 Date 的一个实例。
|
||||
</pre>
|
||||
|
||||
<p><strong>示例 2：</strong></p>
|
||||
|
||||
<pre>
|
||||
<b>输入：</b>func = () => { class Animal {}; class Dog extends Animal {}; return checkIfInstance(new Dog(), Animal); }
|
||||
<b>输出：</b>true
|
||||
<strong>解释：</strong>
|
||||
class Animal {};
|
||||
class Dog extends Animal {};
|
||||
checkIfInstance(new Dog(), Animal); // true
|
||||
|
||||
Dog 是 Animal 的子类。因此，Dog 对象同时是 Dog 和 Animal 的实例。</pre>
|
||||
|
||||
<p><strong>示例 3：</strong></p>
|
||||
|
||||
<pre>
|
||||
<b>输入：</b>func = () => checkIfInstance(Date, Date)
|
||||
<b>输出：</b>false
|
||||
<strong>解释：</strong>日期的构造函数在逻辑上不能是其自身的实例。
|
||||
</pre>
|
||||
|
||||
<p><strong>示例 4：</strong></p>
|
||||
|
||||
<pre>
|
||||
<b>输入：</b>func = () => checkIfInstance(5, Number)
|
||||
<b>输出：</b>true
|
||||
<strong>解释：</strong>5 是一个 Number。注意，"instanceof" 关键字将返回 false。</pre>
|
||||
@ -0,0 +1,55 @@
|
||||
<p>给你一个长度为 <code>n</code> 的整数数组 <code>nums</code> ，请你求出每个长度为 <code>k</code> 的子数组的 <b>美丽值</b> 。</p>
|
||||
|
||||
<p>一个子数组的 <strong>美丽值</strong> 定义为：如果子数组中第 <code>x</code> <strong>小整数</strong> 是 <strong>负数</strong> ，那么美丽值为第 <code>x</code> 小的数，否则美丽值为 <code>0</code> 。</p>
|
||||
|
||||
<p>请你返回一个包含<em> </em><code>n - k + 1</code> 个整数的数组，<strong>依次</strong> 表示数组中从第一个下标开始，每个长度为 <code>k</code> 的子数组的<strong> 美丽值</strong> 。</p>
|
||||
|
||||
<ul>
|
||||
<li>
|
||||
<p>子数组指的是数组中一段连续 <strong>非空</strong> 的元素序列。</p>
|
||||
</li>
|
||||
</ul>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>示例 1：</strong></p>
|
||||
|
||||
<pre><b>输入：</b>nums = [1,-1,-3,-2,3], k = 3, x = 2
|
||||
<b>输出：</b>[-1,-2,-2]
|
||||
<b>解释：</b>总共有 3 个 k = 3 的子数组。
|
||||
第一个子数组是 <code>[1, -1, -3]</code> ，第二小的数是负数 -1 。
|
||||
第二个子数组是 <code>[-1, -3, -2]</code> ，第二小的数是负数 -2 。
|
||||
第三个子数组是 <code>[-3, -2, 3] ，第二小的数是负数 -2 。</code></pre>
|
||||
|
||||
<p><strong>示例 2：</strong></p>
|
||||
|
||||
<pre><b>输入：</b>nums = [-1,-2,-3,-4,-5], k = 2, x = 2
|
||||
<b>输出：</b>[-1,-2,-3,-4]
|
||||
<b>解释：</b>总共有 4 个 k = 2 的子数组。
|
||||
<code>[-1, -2] 中第二小的数是负数 -1 。</code>
|
||||
<code>[-2, -3] 中第二小的数是负数 -2 。</code>
|
||||
<code>[-3, -4] 中第二小的数是负数 -3 。</code>
|
||||
<code>[-4, -5] 中第二小的数是负数 -4 。</code></pre>
|
||||
|
||||
<p><strong>示例 3：</strong></p>
|
||||
|
||||
<pre><b>输入：</b>nums = [-3,1,2,-3,0,-3], k = 2, x = 1
|
||||
<b>输出：</b>[-3,0,-3,-3,-3]
|
||||
<b>解释：</b>总共有 5 个 k = 2 的子数组。
|
||||
<code>[-3, 1] 中最小的数是负数 -3 。</code>
|
||||
<code>[1, 2] 中最小的数不是负数，所以美丽值为 0 。</code>
|
||||
<code>[2, -3] 中最小的数是负数 -3 。</code>
|
||||
<code>[-3, 0] 中最小的数是负数 -3 。</code>
|
||||
<code>[0, -3] 中最小的数是负数 -3 。</code></pre>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>提示：</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>n == nums.length </code></li>
|
||||
<li><code>1 <= n <= 10<sup>5</sup></code></li>
|
||||
<li><code>1 <= k <= n</code></li>
|
||||
<li><code>1 <= x <= k </code></li>
|
||||
<li><code>-50 <= nums[i] <= 50 </code></li>
|
||||
</ul>
|
||||
@ -0,0 +1,37 @@
|
||||
<p>请你编写一个生成器函数，并返回一个可以生成 <strong>斐波那契数列</strong> 的生成器对象。</p>
|
||||
|
||||
<p><strong>斐波那契数列</strong> 的递推公式为 <code>X<sub>n</sub> = X<sub>n-1</sub> + X<sub>n-2</sub></code> 。</p>
|
||||
|
||||
<p>这个数列的前几个数字是 <code>0, 1, 1, 2, 3, 5, 8, 13</code> 。</p>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>示例 1：</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>输入：</strong>callCount = 5
|
||||
<b>输出：</b>[0,1,1,2,3]
|
||||
<strong>解释：</strong>
|
||||
const gen = fibGenerator();
|
||||
gen.next().value; // 0
|
||||
gen.next().value; // 1
|
||||
gen.next().value; // 1
|
||||
gen.next().value; // 2
|
||||
gen.next().value; // 3
|
||||
</pre>
|
||||
|
||||
<p><strong>示例 2：</strong></p>
|
||||
|
||||
<pre>
|
||||
<b>输入：</b>callCount = 0
|
||||
<strong>输出：</strong>[]
|
||||
<b>解释：</b>gen.next() 永远不会被调用，所以什么也不会输出
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>提示：</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>0 <= callCount <= 50</code></li>
|
||||
</ul>
|
||||
32
leetcode-cn/problem (Chinese)/睡眠函数 [sleep].html
Normal file
32
leetcode-cn/problem (Chinese)/睡眠函数 [sleep].html
Normal file
@ -0,0 +1,32 @@
|
||||
<p>请你编写一个异步函数，它接收一个正整数参数 <code>millis</code> ，并休眠这么多毫秒。要求此函数可以解析任何值。</p>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><b>示例 1：</b></p>
|
||||
|
||||
<pre>
|
||||
<b>输入：</b>millis = 100
|
||||
<b>输出：</b>100
|
||||
<b>解释：</b>
|
||||
在 100ms 后此异步函数执行完时返回一个 Promise 对象
|
||||
let t = Date.now();
|
||||
sleep(100).then(() => {
|
||||
console.log(Date.now() - t); // 100
|
||||
});
|
||||
</pre>
|
||||
|
||||
<p><b>示例 2：</b></p>
|
||||
|
||||
<pre>
|
||||
<b>输入：</b>millis = 200
|
||||
<b>输出：</b>200
|
||||
<b>解释：</b>在 200ms 后函数执行完时返回一个 Promise 对象
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><b>提示：</b></p>
|
||||
|
||||
<ul>
|
||||
<li><code>1 <= millis <= 1000</code></li>
|
||||
</ul>
|
||||
58
leetcode-cn/problem (Chinese)/蜗牛排序 [snail-traversal].html
Normal file
58
leetcode-cn/problem (Chinese)/蜗牛排序 [snail-traversal].html
Normal file
@ -0,0 +1,58 @@
|
||||
<p>请你编写一段代码为所有数组实现 <code>snail(rowsCount，colsCount)</code> 方法，该方法将 1D 数组转换为以蜗牛排序的模式的 2D 数组。无效的输入值应该输出一个空数组。当 <code>rowsCount * colsCount !==</code><code>nums.length</code> 时。这个输入被认为是无效的。</p>
|
||||
|
||||
<p>蜗牛排序从左上角的单元格开始，从当前数组的第一个值开始。然后，它从上到下遍历第一列，接着移动到右边的下一列，并从下到上遍历它。将这种模式持续下去，每列交替变换遍历方向，直到覆盖整个数组。例如，当给定输入数组 <code>[19, 10, 3, 7, 9, 8, 5, 2, 1, 17, 16, 14, 12, 18, 6, 13, 11, 20, 4, 15]</code> ，当 <code>rowsCount = 5</code> 且 <code>colsCount = 4</code> 时，需要输出矩阵如下图所示。注意，矩阵沿箭头方向对应于原数组中数字的顺序</p>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><img alt="Traversal Diagram" src="https://assets.leetcode.com/uploads/2023/04/10/screen-shot-2023-04-10-at-100006-pm.png" style="width: 275px; height: 343px;" /></p>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><b>示例 1：</b></p>
|
||||
|
||||
<pre>
|
||||
<b>输入：</b>
|
||||
nums = [19, 10, 3, 7, 9, 8, 5, 2, 1, 17, 16, 14, 12, 18, 6, 13, 11, 20, 4, 15]
|
||||
rowsCount = 5
|
||||
colsCount = 4
|
||||
<b>输出：</b>
|
||||
[
|
||||
[19,17,16,15],
|
||||
[10,1,14,4],
|
||||
[3,2,12,20],
|
||||
[7,5,18,11],
|
||||
[9,8,6,13]
|
||||
]
|
||||
</pre>
|
||||
|
||||
<p><b>示例 2：</b></p>
|
||||
|
||||
<pre>
|
||||
<b>输入：</b>
|
||||
nums = [1,2,3,4]
|
||||
rowsCount = 1
|
||||
colsCount = 4
|
||||
<b>输出：</b>[[1, 2, 3, 4]]
|
||||
</pre>
|
||||
|
||||
<p><b>示例 3：</b></p>
|
||||
|
||||
<pre>
|
||||
<b>输入：</b>
|
||||
nums = [1,3]
|
||||
rowsCount = 2
|
||||
colsCount = 2
|
||||
<b>输出：</b>[]
|
||||
<strong>Explanation:</strong> 2 * 2 = 4, 且原数组 [1,3] 的长度为 2; 所以，输入是无效的。
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><b>提示：</b></p>
|
||||
|
||||
<ul>
|
||||
<li><code>0 <= nums.length <= 250</code></li>
|
||||
<li><code>1 <= nums[i] <= 1000</code></li>
|
||||
<li><code>1 <= rowsCount <= 250</code></li>
|
||||
<li><code>1 <= colsCount <= 250</code></li>
|
||||
</ul>
|
||||
26
leetcode-cn/problem (Chinese)/补给马车 [hqCnmP].md
Normal file
26
leetcode-cn/problem (Chinese)/补给马车 [hqCnmP].md
Normal file
@ -0,0 +1,26 @@
|
||||
远征队即将开启未知的冒险之旅，不过在此之前，将对补给车队进行最后的检查。`supplies[i]` 表示编号为 `i` 的补给马车装载的物资数量。
|
||||
考虑到车队过长容易被野兽偷袭，他们决定将车队的长度变为原来的一半（向下取整），计划为：
|
||||
- 找出车队中 **物资之和最小** 两辆 **相邻** 马车，将它们车辆的物资整合为一辆。若存在多组物资之和相同的马车，则取编号最小的两辆马车进行整合；
|
||||
- 重复上述操作直到车队长度符合要求。
|
||||
|
||||
请返回车队长度符合要求后，物资的分布情况。
|
||||
|
||||
**示例 1：**
|
||||
>输入：`supplies = [7,3,6,1,8]`
|
||||
>
|
||||
>输出：`[10,15]`
|
||||
>
|
||||
>解释：
|
||||
> 第 1 次合并，符合条件的两辆马车为 6,1，合并后的车队为 [7,3,7,8]；
|
||||
> 第 2 次合并，符合条件的两辆马车为 (7,3) 和 (3,7)，取编号最小的 (7,3)，合并后的车队为 [10,7,8]；
|
||||
> 第 3 次合并，符合条件的两辆马车为 7,8，合并后的车队为 [10,15]；
|
||||
>返回 `[10,15]`
|
||||
|
||||
**示例 2：**
|
||||
>输入：`supplies = [1,3,1,5]`
|
||||
>
|
||||
>输出：`[5,5]`
|
||||
|
||||
**解释：**
|
||||
- `2 <= supplies.length <= 1000`
|
||||
- `1 <= supplies[i] <= 1000`
|
||||
35
leetcode-cn/problem (Chinese)/计数器 [counter].html
Normal file
35
leetcode-cn/problem (Chinese)/计数器 [counter].html
Normal file
@ -0,0 +1,35 @@
|
||||
<p>请你编写并返回一个 <strong>计数器 </strong>函数，它接收一个整型参数 n 。这个 <strong>计数器 </strong>函数最初返回 n，每次调用它时返回前一个值加 1 的值 ( <code>n</code> , <code>n + 1</code> , <code>n + 2</code> ，等等)。</p>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>示例 1：</strong></p>
|
||||
|
||||
<pre>
|
||||
<b>输入：</b>
|
||||
n = 10
|
||||
["call","call","call"]
|
||||
<b>输出：</b>[10,11,12]
|
||||
<strong>解释：
|
||||
</strong>counter() = 10 // 第一次调用 counter()，返回 n。
|
||||
counter() = 11 // 返回上次调用的值加 1。
|
||||
counter() = 12 // 返回上次调用的值加 1。
|
||||
</pre>
|
||||
|
||||
<p><strong>示例 2：</strong></p>
|
||||
|
||||
<pre>
|
||||
<b>输入：</b>
|
||||
n = -2
|
||||
["call","call","call","call","call"]
|
||||
<b>输出：</b>[-2,-1,0,1,2]
|
||||
<b>解释：</b>counter() 最初返回 -2。然后在每个后续调用后增加 1。
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>提示：</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>-1000<sup> </sup><= n <= 1000</code></li>
|
||||
<li><code>最多对 counter() 进行 1000 次调用</code></li>
|
||||
</ul>
|
||||
@ -0,0 +1,29 @@
|
||||
<p>给你一个正整数 <code>arrivalTime</code> 表示列车正点到站的时间（单位：小时），另给你一个正整数 <code>delayedTime</code> 表示列车延误的小时数。</p>
|
||||
|
||||
<p>返回列车实际到站的时间。</p>
|
||||
|
||||
<p>注意，该问题中的时间采用 24 小时制。</p>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>示例 1：</strong></p>
|
||||
|
||||
<pre><strong>输入：</strong>arrivalTime = 15, delayedTime = 5
|
||||
<strong>输出：</strong>20
|
||||
<strong>解释：</strong>列车正点到站时间是 15:00 ，延误 5 小时，所以列车实际到站的时间是 15 + 5 = 20（20:00）。
|
||||
</pre>
|
||||
|
||||
<p><strong>示例 2：</strong></p>
|
||||
|
||||
<pre><strong>输入：</strong>arrivalTime = 13, delayedTime = 11
|
||||
<strong>输出：</strong>0
|
||||
<strong>解释：</strong>列车正点到站时间是 13:00 ，延误 11 小时，所以列车实际到站的时间是 13 + 11 = 24（在 24 小时制中表示为 00:00 ，所以返回 0）。</pre>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>提示：</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>1 <= arrivaltime < 24</code></li>
|
||||
<li><code>1 <= delayedTime <= 24</code></li>
|
||||
</ul>
|
||||
58
leetcode-cn/problem (Chinese)/记忆函数 II [memoize-ii].html
Normal file
58
leetcode-cn/problem (Chinese)/记忆函数 II [memoize-ii].html
Normal file
@ -0,0 +1,58 @@
|
||||
<p>请你编写一个函数，它接收一个函数参数 <code>fn</code>，并返回该函数的 <strong>记忆化</strong> 后的结果。</p>
|
||||
|
||||
<p><strong>记忆函数</strong> 是一个对于相同的输入永远不会被调用两次的函数。相反，它将返回一个缓存值。</p>
|
||||
|
||||
<p><code>fn</code> 可以是任何函数，对于它接受什么类型的值没有限制。如果输入为 <code>===</code>，则认为输入相同。</p>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>示例 1：</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>输入：</strong>
|
||||
getInputs = () => [[2,2],[2,2],[1,2]]
|
||||
fn = function (a, b) { return a + b; }
|
||||
<b>输出：</b>[{"val":4,"calls":1},{"val":4,"calls":1},{"val":3,"calls":2}]
|
||||
<strong>解释：</strong>
|
||||
const inputs = getInputs();
|
||||
const memoized = memoize(fn);
|
||||
for (const arr of inputs) {
|
||||
memoized(...arr);
|
||||
}
|
||||
|
||||
对于参数为 (2, 2) 的输入: 2 + 2 = 4，需要调用 fn() 。
|
||||
对于参数为 (2, 2) 的输入: 2 + 2 = 4，这些输入此前调用过，因此不需要调用 fn() 。
|
||||
对于参数为 (1, 2) 的输入: 1 + 2 = 3，需要再次调用 fn()，总调用数为 2 。
|
||||
</pre>
|
||||
|
||||
<p><strong>示例 2：</strong></p>
|
||||
|
||||
<pre>
|
||||
<b>输入：</b>
|
||||
getInputs = () => [[{},{}],[{},{}],[{},{}]]
|
||||
fn = function (a, b) { return a + b; }
|
||||
<b>输出：</b>[{"val":{},"calls":1},{"val":{},"calls":2},{"val":{},"calls":3}]
|
||||
<strong>解释：</strong>
|
||||
合并两个空对象总是会得到一个空对象。因为缓存命中，所以只有 1 次对 fn() 的调用，尽管这些对象之间没有一个是相同的（===）。
|
||||
</pre>
|
||||
|
||||
<p><strong>示例 3：</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>输入：</strong>
|
||||
getInputs = () => { const o = {}; return [[o,o],[o,o],[o,o]]; }
|
||||
fn = function (a, b) { return ({...a, ...b}); }
|
||||
<b>输出：</b>[{"val":{},"calls":1},{"val":{},"calls":1},{"val":{},"calls":1}]
|
||||
<strong>解释：</strong>
|
||||
合并两个空对象总是会得到一个空对象。第 2 和第 3 个函数调用导致缓存命中。这是因为传入的每个对象都是相同的。
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>提示：</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>1 <= inputs.length <= 10<sup>5</sup></code></li>
|
||||
<li><code>0 <= inputs.flat().length <= 10<sup>5</sup></code></li>
|
||||
<li><code>inputs[i][j] != NaN</code></li>
|
||||
</ul>
|
||||
82
leetcode-cn/problem (Chinese)/记忆函数 [memoize].html
Normal file
82
leetcode-cn/problem (Chinese)/记忆函数 [memoize].html
Normal file
@ -0,0 +1,82 @@
|
||||
<p>请你编写一个函数，它接收另一个函数作为输入，并返回该函数的 <strong>记忆化</strong> 后的结果。</p>
|
||||
|
||||
<p><strong>记忆函数</strong> 是一个对于相同的输入永远不会被调用两次的函数。相反，它将返回一个缓存值。</p>
|
||||
|
||||
<p>你可以假设有 <strong>3</strong> 个可能的输入函数：<code>sum</code> 、<code>fib</code> 和 <code>factorial</code> 。</p>
|
||||
|
||||
<ul>
|
||||
<li> <code>sum</code> 接收两个整型参数 <code>a</code> 和 <code>b</code> ，并返回 <code>a + b</code> 。</li>
|
||||
<li> <code>fib</code> 接收一个整型参数 <code>n</code> ，如果 <code>n <= 1</code> 则返回 <code>1</code>，否则返回 <code>fib (n - 1) + fib (n - 2)</code>。</li>
|
||||
<li> <code>factorial</code> 接收一个整型参数 <code>n</code> ，如果 <code>n <= 1</code> 则返回 <code>1</code> ，否则返回 <code>factorial(n - 1) * n</code> 。</li>
|
||||
</ul>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>示例 1：</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>输入：</strong>
|
||||
"sum"
|
||||
["call","call","getCallCount","call","getCallCount"]
|
||||
[[2,2],[2,2],[],[1,2],[]]
|
||||
<strong>输出：</strong>
|
||||
[4,4,1,3,2]
|
||||
|
||||
<strong>解释：</strong>
|
||||
const sum = (a, b) => a + b;
|
||||
const memoizedSum = memoize(sum);
|
||||
memoizedSum (2, 2);// 返回 4。sum() 被调用，因为之前没有使用参数 (2, 2) 调用过。
|
||||
memoizedSum (2, 2);// 返回 4。没有调用 sum()，因为前面有相同的输入。
|
||||
//总调用数： 1
|
||||
memoizedSum(1、2);// 返回 3。sum() 被调用，因为之前没有使用参数 (1, 2) 调用过。
|
||||
//总调用数： 2
|
||||
</pre>
|
||||
|
||||
<p><strong>示例 2：</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>输入：
|
||||
</strong>"factorial"
|
||||
["call","call","call","getCallCount","call","getCallCount"]
|
||||
[[2],[3],[2],[],[3],[]]
|
||||
<strong>输出：</strong>
|
||||
[2,6,2,2,6,2]
|
||||
|
||||
<strong>解释：</strong>
|
||||
const factorial = (n) => (n <= 1) ? 1 : (n * factorial(n - 1));
|
||||
const memoFactorial = memoize(factorial);
|
||||
memoFactorial(2); // 返回 2。
|
||||
memoFactorial(3); // 返回 6。
|
||||
memoFactorial(2); // 返回 2。 没有调用 factorial()，因为前面有相同的输入。
|
||||
// 总调用数：2
|
||||
memoFactorial(3); // 返回 6。 没有调用 factorial()，因为前面有相同的输入。
|
||||
// 总调用数：2
|
||||
</pre>
|
||||
|
||||
<p><strong>示例 3：</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>输入：
|
||||
</strong>"fib"
|
||||
["call","getCallCount"]
|
||||
[[5],[]]
|
||||
<strong>输出：</strong>
|
||||
[8,1]
|
||||
|
||||
<strong>解释：
|
||||
</strong>fib(5) = 8
|
||||
// 总调用数：1
|
||||
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>Constraints:</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>0 <= a, b <= 10<sup>5</sup></code></li>
|
||||
<li><code>1 <= n <= 10</code></li>
|
||||
<li><code>at most 10<sup>5</sup> function calls</code></li>
|
||||
<li><code>at most 10<sup>5</sup> attempts to access callCount</code></li>
|
||||
<li><code>input function is sum, fib, or factorial</code></li>
|
||||
</ul>
|
||||
@ -0,0 +1,44 @@
|
||||
<p>给你一个有 <code>n</code> 个节点的 <strong>有向带权</strong> 图，节点编号为 <code>0</code> 到 <code>n - 1</code> 。图中的初始边用数组 <code>edges</code> 表示，其中 <code>edges[i] = [from<sub>i</sub>, to<sub>i</sub>, edgeCost<sub>i</sub>]</code> 表示从 <code>from<sub>i</sub></code> 到 <code>to<sub>i</sub></code> 有一条代价为 <code>edgeCost<sub>i</sub></code> 的边。</p>
|
||||
|
||||
<p>请你实现一个 <code>Graph</code> 类：</p>
|
||||
|
||||
<ul>
|
||||
<li><code>Graph(int n, int[][] edges)</code> 初始化图有 <code>n</code> 个节点，并输入初始边。</li>
|
||||
<li><code>addEdge(int[] edge)</code> 向边集中添加一条边，其中<strong> </strong><code>edge = [from, to, edgeCost]</code> 。数据保证添加这条边之前对应的两个节点之间没有有向边。</li>
|
||||
<li><code>int shortestPath(int node1, int node2)</code> 返回从节点 <code>node1</code> 到 <code>node2</code> 的路径<strong> 最小</strong> 代价。如果路径不存在，返回 <code>-1</code> 。一条路径的代价是路径中所有边代价之和。</li>
|
||||
</ul>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>示例 1：</strong></p>
|
||||
|
||||
<p><img alt="" src="https://assets.leetcode.com/uploads/2023/01/11/graph3drawio-2.png" style="width: 621px; height: 191px;"></p>
|
||||
|
||||
<pre><strong>输入：</strong>
|
||||
["Graph", "shortestPath", "shortestPath", "addEdge", "shortestPath"]
|
||||
[[4, [[0, 2, 5], [0, 1, 2], [1, 2, 1], [3, 0, 3]]], [3, 2], [0, 3], [[1, 3, 4]], [0, 3]]
|
||||
<b>输出：</b>
|
||||
[null, 6, -1, null, 6]
|
||||
|
||||
<strong>解释：</strong>
|
||||
Graph g = new Graph(4, [[0, 2, 5], [0, 1, 2], [1, 2, 1], [3, 0, 3]]);
|
||||
g.shortestPath(3, 2); // 返回 6 。从 3 到 2 的最短路径如第一幅图所示：3 -> 0 -> 1 -> 2 ，总代价为 3 + 2 + 1 = 6 。
|
||||
g.shortestPath(0, 3); // 返回 -1 。没有从 0 到 3 的路径。
|
||||
g.addEdge([1, 3, 4]); // 添加一条节点 1 到节点 3 的边，得到第二幅图。
|
||||
g.shortestPath(0, 3); // 返回 6 。从 0 到 3 的最短路径为 0 -> 1 -> 3 ，总代价为 2 + 4 = 6 。
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>提示：</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>1 <= n <= 100</code></li>
|
||||
<li><code>0 <= edges.length <= n * (n - 1)</code></li>
|
||||
<li><code>edges[i].length == edge.length == 3</code></li>
|
||||
<li><code>0 <= from<sub>i</sub>, to<sub>i</sub>, from, to, node1, node2 <= n - 1</code></li>
|
||||
<li><code>1 <= edgeCost<sub>i</sub>, edgeCost <= 10<sup>6</sup></code></li>
|
||||
<li>图中任何时候都不会有重边和自环。</li>
|
||||
<li>调用 <code>addEdge</code> 至多 <code>100</code> 次。</li>
|
||||
<li>调用 <code>shortestPath</code> 至多 <code>100</code> 次。</li>
|
||||
</ul>
|
||||
@ -0,0 +1,137 @@
|
||||
<p>有时候你会有一个长时间运行的任务，并且你可能希望在它完成之前取消它。为了实现这个目标，请你编写一个名为 <code>cancellable</code> 的函数，它接收一个生成器对象，并返回一个包含两个值的数组：一个 <strong>取消函数</strong> 和一个 <strong>promise</strong> 对象。</p>
|
||||
|
||||
<p>你可以假设生成器函数只会生成 promise 对象。你的函数负责将 promise 对象解析的值传回生成器。如果 promise 被拒绝，你的函数应将该错误抛回给生成器。</p>
|
||||
|
||||
<p>如果在生成器完成之前调用了取消回调函数，则你的函数应该将错误抛回给生成器。该错误应该是字符串 <code>"Cancelled"</code>（而不是一个 <code>Error</code> 对象）。如果错误被捕获，则返回的 promise 应该解析为下一个生成或返回的值。否则，promise 应该被拒绝并抛出该错误。不应执行任何其他代码。</p>
|
||||
|
||||
<p>当生成器完成时，您的函数返回的 promise 应该解析为生成器返回的值。但是，如果生成器抛出错误，则返回的 promise 应该拒绝并抛出该错误。</p>
|
||||
|
||||
<p>下面是您的代码应如何使用的示例：</p>
|
||||
|
||||
<pre>
|
||||
function* tasks() {
|
||||
const val = yield new Promise(resolve => resolve(2 + 2));
|
||||
yield new Promise(resolve => setTimeout(resolve, 100));
|
||||
return val + 1; // calculation shouldn't be done.
|
||||
}
|
||||
const [cancel, promise] = cancellable(tasks());
|
||||
setTimeout(cancel, 50);
|
||||
promise.catch(console.log); // logs "Cancelled" at t=50ms
|
||||
</pre>
|
||||
|
||||
<p>如果相反， <code>cancel()</code> 没有被调用或者在 <code>t=100ms</code> 之后才被调用，那么 Promise 应被解析为 <code>5</code> 。</p>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>示例 1：</strong></p>
|
||||
|
||||
<pre>
|
||||
<b>输入：</b>
|
||||
generatorFunction = function*() {
|
||||
return 42;
|
||||
}
|
||||
cancelledAt = 100
|
||||
<b>输出：</b>{"resolved": 42}
|
||||
<strong>解释：</strong>
|
||||
const generator = generatorFunction();
|
||||
const [cancel, promise] = cancellable(generator);
|
||||
setTimeout(cancel, 100);
|
||||
promise.then(console.log); // 在 t=0ms 解析为 42
|
||||
|
||||
该生成器立即生成 42 并完成。因此，返回的 promise 立即解析为 42。请注意，取消已经完成的生成器没有任何作用。
|
||||
</pre>
|
||||
|
||||
<p><strong>示例 2：</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>输入：</strong>
|
||||
generatorFunction = function*() {
|
||||
const msg = yield new Promise(res => res("Hello"));
|
||||
throw `Error: ${msg}`;
|
||||
}
|
||||
cancelledAt = null
|
||||
<b>输出：</b>{"rejected": "Error: Hello"}
|
||||
<strong>解释：</strong>
|
||||
一个 Promise 被生成。该函数通过等待 promise 解析并将解析后的值传回生成器来处理它。然后抛出一个错误，这会导致 promise 被同样抛出的错误拒绝。
|
||||
</pre>
|
||||
|
||||
<p><strong>示例 3：</strong></p>
|
||||
|
||||
<pre>
|
||||
<b>输入：</b>
|
||||
generatorFunction = function*() {
|
||||
yield new Promise(res => setTimeout(res, 200));
|
||||
return "Success";
|
||||
}
|
||||
cancelledAt = 100
|
||||
<b>输出：</b>{"rejected": "Cancelled"}
|
||||
<strong>解释：</strong>
|
||||
当函数等待被生成的 promise 解析时，cancel() 被调用。这会导致一个错误消息被发送回生成器。由于这个错误没有被捕获，返回的 promise 会因为这个错误而被拒绝。
|
||||
</pre>
|
||||
|
||||
<p><strong>示例 4：</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>输入：</strong>
|
||||
generatorFunction = function*() {
|
||||
let result = 0;
|
||||
yield new Promise(res => setTimeout(res, 100));
|
||||
result += yield new Promise(res => res(1));
|
||||
yield new Promise(res => setTimeout(res, 100));
|
||||
result += yield new Promise(res => res(1));
|
||||
return result;
|
||||
}
|
||||
cancelledAt = null
|
||||
<b>输出：</b>{"resolved": 2}
|
||||
<strong>解释：</strong>
|
||||
生成器生成了 4 个 promise 。其中两个 promise 的值被添加到结果中。200ms 后，生成器以值 2 完成，该值被返回的 promise 解析。
|
||||
</pre>
|
||||
|
||||
<p><strong>示例 5：</strong></p>
|
||||
|
||||
<pre>
|
||||
<b>输入：</b>
|
||||
generatorFunction = function*() {
|
||||
let result = 0;
|
||||
try {
|
||||
yield new Promise(res => setTimeout(res, 100));
|
||||
result += yield new Promise(res => res(1));
|
||||
yield new Promise(res => setTimeout(res, 100));
|
||||
result += yield new Promise(res => res(1));
|
||||
} catch(e) {
|
||||
return result;
|
||||
}
|
||||
return result;
|
||||
}
|
||||
cancelledAt = 150
|
||||
<b>输出：</b>{"resolved": 1}
|
||||
<strong>解释：</strong>
|
||||
前两个生成的 promise 解析并导致结果递增。然而，在 t=150ms 时，生成器被取消了。发送给生成器的错误被捕获，结果被返回并最终由返回的 promise 解析。
|
||||
</pre>
|
||||
|
||||
<p><strong>示例 6：</strong></p>
|
||||
|
||||
<pre>
|
||||
<b>输入：</b>
|
||||
generatorFunction = function*() {
|
||||
try {
|
||||
yield new Promise((resolve, reject) => reject("Promise Rejected"));
|
||||
} catch(e) {
|
||||
let a = yield new Promise(resolve => resolve(2));
|
||||
let b = yield new Promise(resolve => resolve(2));
|
||||
return a + b;
|
||||
};
|
||||
}
|
||||
cancelledAt = null
|
||||
<b>输出：</b>{"resolved": 4}
|
||||
<strong>解释：</strong>
|
||||
第一个生成的 promise 立即被拒绝。该错误被捕获。因为生成器没有被取消，执行继续像往常一样。最终解析为 2 + 2 = 4。</pre>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>提示：</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>cancelledAt == null or 0 <= cancelledAt <= 1000</code></li>
|
||||
<li><code>generatorFunction 返回一个生成器对象</code></li>
|
||||
</ul>
|
||||
@ -0,0 +1,44 @@
|
||||
<p>编写一个函数，这个函数接收一个整数数组 <code>arr</code> 和一个映射函数 <code>fn</code> ，通过该映射函数返回一个新的数组。</p>
|
||||
|
||||
<p>返回数组的创建语句应为 <code>returnedArray[i] = fn(arr[i], i)</code> 。</p>
|
||||
|
||||
<p>请你在不使用内置方法 <code>Array.map</code> 的前提下解决这个问题。</p>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong class="example">示例 1:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>输入：</strong>arr = [1,2,3], fn = function plusone(n) { return n + 1; }
|
||||
<strong>输出：</strong>[2,3,4]
|
||||
<strong>解释： </strong>
|
||||
const newArray = map(arr, plusone); // [2,3,4]
|
||||
此映射函数返回值是将数组中每个元素的值加 1。
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">示例</strong><strong class="example"> 2:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>输入：</strong>arr = [1,2,3], fn = function plusI(n, i) { return n + i; }
|
||||
<strong>输出：</strong>[1,3,5]
|
||||
<strong>解释：</strong>此映射函数返回值根据输入数组索引增加每个值。
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">示例 3:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>输入：</strong>arr = [10,20,30], fn = function constant() { return 42; }
|
||||
<strong>输出：</strong>[42,42,42]
|
||||
<strong>解释：</strong>此映射函数返回值恒为 42。
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>提示：</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>0 <= arr.length <= 1000</code></li>
|
||||
<li><code><font face="monospace">-10<sup>9</sup> <= arr[i] <= 10<sup>9</sup></font></code></li>
|
||||
<li><font face="monospace"><code>fn 返回一个数</code></font></li>
|
||||
</ul>
|
||||
<span style="display:block"><span style="height:0px"><span style="position:absolute"></span></span></span>
|
||||
@ -0,0 +1,44 @@
|
||||
<p>请你编写一个函数，该函数接受一个整数数组参数 <code>arr</code> 和一个过滤函数 <code>fn</code>，并返回一个过滤后元素数量较少或元素数量相等的新数组。</p>
|
||||
|
||||
<p>返回的数组应该只包含通过过滤函数 <code>fn(arr[i]， i)</code> 计算后为真值的元素。</p>
|
||||
|
||||
<p>请你在不使用内置函数 <code>Array.filter</code> 的前提下解决该问题。</p>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong class="example">示例 1：</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>输入：</strong>arr = [0,10,20,30], fn = function greaterThan10(n) { return n > 10; }
|
||||
<b>输出：</b> [20,30]
|
||||
<b>解释：</b>
|
||||
const newArray = filter(arr, fn); // [20, 30]
|
||||
过滤函数过滤掉不大于 10 的值</pre>
|
||||
|
||||
<p><strong class="example">示例 2：</strong></p>
|
||||
|
||||
<pre>
|
||||
<b>输入：</b>arr = [1,2,3], fn = function firstIndex(n, i) { return i === 0; }
|
||||
<b>输出：</b>[1]
|
||||
<strong>解释：</strong>
|
||||
过滤函数 fn 也可以接受每个元素的索引
|
||||
在这种情况下，过滤函数删除索引不为 0 的元素
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">示例 3：</strong></p>
|
||||
|
||||
<pre>
|
||||
<b>输入：</b>arr = [-2,-1,0,1,2], fn = function plusOne(n) { return n + 1 }
|
||||
<b>输出：</b>[-2,0,1,2]
|
||||
<strong>解释：</strong>
|
||||
像 0 这样的假值应被过滤掉
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>提示：</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>0 <= arr.length <= 1000</code></li>
|
||||
<li><code><font face="monospace">-10<sup>9</sup> <= arr[i] <= 10<sup>9</sup></font></code></li>
|
||||
</ul>
|
||||
32
leetcode-cn/problem (Chinese)/魔法棋盘 [1ybDKD].md
Normal file
32
leetcode-cn/problem (Chinese)/魔法棋盘 [1ybDKD].md
Normal file
@ -0,0 +1,32 @@
|
||||
在大小为 `n * m` 的棋盘中，有两种不同的棋子：黑色，红色。当两颗颜色不同的棋子同时满足以下两种情况时，将会产生魔法共鸣：
|
||||
- 两颗异色棋子在同一行或者同一列
|
||||
- 两颗异色棋子之间恰好只有一颗棋子
|
||||
|
||||
由于棋盘上被施加了魔法禁制，棋盘上的部分格子变成问号。`chessboard[i][j]` 表示棋盘第 `i` 行 `j` 列的状态：
|
||||
- 若为 `.` ，表示当前格子确定为空
|
||||
- 若为 `B` ，表示当前格子确定为 黑棋
|
||||
- 若为 `R` ，表示当前格子确定为 红棋
|
||||
- 若为 `?` ，表示当前格子待定
|
||||
|
||||
现在，探险家小扣的任务是确定所有问号位置的状态（留空/放黑棋/放红棋），使最终的棋盘上，任意两颗棋子间都 **无法** 产生共鸣。请返回可以满足上述条件的放置方案数量。
|
||||
|
||||
**示例1：**
|
||||
> 输入：`n = 3, m = 3, chessboard = ["..R","..B","?R?"]`
|
||||
>
|
||||
> 输出：`5`
|
||||
>
|
||||
> 解释：给定的棋盘如图：
|
||||
>{:height=150px}
|
||||
> 所有符合题意的最终局面如图：
|
||||
>{:height=150px}
|
||||
|
||||
**示例2：**
|
||||
> 输入：`n = 3, m = 3, chessboard = ["?R?","B?B","?R?"]`
|
||||
>
|
||||
> 输出：`105`
|
||||
|
||||
**提示：**
|
||||
- `n == chessboard.length`
|
||||
- `m == chessboard[i].length`
|
||||
- `1 <= n*m <= 30`
|
||||
- `chessboard` 中仅包含 `"."、"B"、"R"、"?"`
|
||||
@ -0,0 +1,71 @@
|
||||
<p>Given an array of asyncronous functions <code>functions</code> and a <strong>pool limit</strong> <code>n</code>, return an asyncronous function <code>promisePool</code>. It should return a promise that resolves when all the input functions resolve.</p>
|
||||
|
||||
<p><b>Pool limit</b> is defined as the maximum number promises that can be pending at once. <code>promisePool</code> should begin execution of as many functions as possible and continue executing new functions when old promises resolve. <code>promisePool</code> should execute <code>functions[i]</code> then <code>functions[i + 1]</code> then <code>functions[i + 2]</code>, etc. When the last promise resolves, <code>promisePool</code> should also resolve.</p>
|
||||
|
||||
<p>For example, if <code>n = 1</code>, <code>promisePool</code> will execute one function at a time in series. However, if <code>n = 2</code>, it first executes two functions. When either of the two functions resolve, a 3rd function should be executed (if available), and so on until there are no functions left to execute.</p>
|
||||
|
||||
<p>You can assume all <code>functions</code> never reject. It is acceptable for <code>promisePool</code> to return a promise that resolves any value.</p>
|
||||
|
||||
<p> </p>
|
||||
<p><strong class="example">Example 1:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong>
|
||||
functions = [
|
||||
() => new Promise(res => setTimeout(res, 300)),
|
||||
() => new Promise(res => setTimeout(res, 400)),
|
||||
() => new Promise(res => setTimeout(res, 200))
|
||||
]
|
||||
n = 2
|
||||
<strong>Output:</strong> [[300,400,500],500]
|
||||
<strong>Explanation:</strong>
|
||||
Three functions are passed in. They sleep for 300ms, 400ms, and 200ms respectively.
|
||||
At t=0, the first 2 functions are executed. The pool size limit of 2 is reached.
|
||||
At t=300, the 1st function resolves, and the 3rd function is executed. Pool size is 2.
|
||||
At t=400, the 2nd function resolves. There is nothing left to execute. Pool size is 1.
|
||||
At t=500, the 3rd function resolves. Pool size is zero so the returned promise also resolves.
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 2:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:
|
||||
</strong>functions = [
|
||||
() => new Promise(res => setTimeout(res, 300)),
|
||||
() => new Promise(res => setTimeout(res, 400)),
|
||||
() => new Promise(res => setTimeout(res, 200))
|
||||
]
|
||||
n = 5
|
||||
<strong>Output:</strong> [[300,400,200],400]
|
||||
<strong>Explanation:</strong>
|
||||
At t=0, all 3 functions are executed. The pool limit of 5 is never met.
|
||||
At t=200, the 3rd function resolves. Pool size is 2.
|
||||
At t=300, the 1st function resolved. Pool size is 1.
|
||||
At t=400, the 2nd function resolves. Pool size is 0, so the returned promise also resolves.
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 3:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong>
|
||||
functions = [
|
||||
() => new Promise(res => setTimeout(res, 300)),
|
||||
() => new Promise(res => setTimeout(res, 400)),
|
||||
() => new Promise(res => setTimeout(res, 200))
|
||||
]
|
||||
n = 1
|
||||
<strong>Output:</strong> [[300,700,900],900]
|
||||
<strong>Explanation:</strong>
|
||||
At t=0, the 1st function is executed. Pool size is 1.
|
||||
At t=300, the 1st function resolves and the 2nd function is executed. Pool size is 1.
|
||||
At t=700, the 2nd function resolves and the 3rd function is executed. Pool size is 1.
|
||||
At t=900, the 3rd function resolves. Pool size is 0 so the returned promise resolves.
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
<p><strong>Constraints:</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>0 <= functions.length <= 10</code></li>
|
||||
<li><code><font face="monospace">1 <= n <= 10</font></code></li>
|
||||
</ul>
|
||||
@ -0,0 +1,45 @@
|
||||
<p>We define the <strong>conversion array</strong> <code>conver</code> of an array <code>arr</code> as follows:</p>
|
||||
|
||||
<ul>
|
||||
<li><code>conver[i] = arr[i] + max(arr[0..i])</code> where <code>max(arr[0..i])</code> is the maximum value of <code>arr[j]</code> over <code>0 <= j <= i</code>.</li>
|
||||
</ul>
|
||||
|
||||
<p>We also define the <strong>score</strong> of an array <code>arr</code> as the sum of the values of the conversion array of <code>arr</code>.</p>
|
||||
|
||||
<p>Given a <strong>0-indexed</strong> integer array <code>nums</code> of length <code>n</code>, return <em>an array </em><code>ans</code><em> of length </em><code>n</code><em> where </em><code>ans[i]</code><em> is the score of the prefix</em> <code>nums[0..i]</code>.</p>
|
||||
|
||||
<p> </p>
|
||||
<p><strong class="example">Example 1:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> nums = [2,3,7,5,10]
|
||||
<strong>Output:</strong> [4,10,24,36,56]
|
||||
<strong>Explanation:</strong>
|
||||
For the prefix [2], the conversion array is [4] hence the score is 4
|
||||
For the prefix [2, 3], the conversion array is [4, 6] hence the score is 10
|
||||
For the prefix [2, 3, 7], the conversion array is [4, 6, 14] hence the score is 24
|
||||
For the prefix [2, 3, 7, 5], the conversion array is [4, 6, 14, 12] hence the score is 36
|
||||
For the prefix [2, 3, 7, 5, 10], the conversion array is [4, 6, 14, 12, 20] hence the score is 56
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 2:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> nums = [1,1,2,4,8,16]
|
||||
<strong>Output:</strong> [2,4,8,16,32,64]
|
||||
<strong>Explanation:</strong>
|
||||
For the prefix [1], the conversion array is [2] hence the score is 2
|
||||
For the prefix [1, 1], the conversion array is [2, 2] hence the score is 4
|
||||
For the prefix [1, 1, 2], the conversion array is [2, 2, 4] hence the score is 8
|
||||
For the prefix [1, 1, 2, 4], the conversion array is [2, 2, 4, 8] hence the score is 16
|
||||
For the prefix [1, 1, 2, 4, 8], the conversion array is [2, 2, 4, 8, 16] hence the score is 32
|
||||
For the prefix [1, 1, 2, 4, 8, 16], the conversion array is [2, 2, 4, 8, 16, 32] hence the score is 64
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
<p><strong>Constraints:</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>1 <= nums.length <= 10<sup>5</sup></code></li>
|
||||
<li><code>1 <= nums[i] <= 10<sup>9</sup></code></li>
|
||||
</ul>
|
||||
@ -0,0 +1,40 @@
|
||||
<p>Given a <code>m x n</code> binary matrix <code>mat</code>, find the <strong>0-indexed</strong> position of the row that contains the <strong>maximum</strong> count of <strong>ones,</strong> and the number of ones in that row.</p>
|
||||
|
||||
<p>In case there are multiple rows that have the maximum count of ones, the row with the <strong>smallest row number</strong> should be selected.</p>
|
||||
|
||||
<p>Return<em> an array containing the index of the row, and the number of ones in it.</em></p>
|
||||
|
||||
<p> </p>
|
||||
<p><strong class="example">Example 1:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> mat = [[0,1],[1,0]]
|
||||
<strong>Output:</strong> [0,1]
|
||||
<strong>Explanation:</strong> Both rows have the same number of 1's. So we return the index of the smaller row, 0, and the maximum count of ones (1<code>)</code>. So, the answer is [0,1].
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 2:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> mat = [[0,0,0],[0,1,1]]
|
||||
<strong>Output:</strong> [1,2]
|
||||
<strong>Explanation:</strong> The row indexed 1 has the maximum count of ones <code>(2)</code>. So we return its index, <code>1</code>, and the count. So, the answer is [1,2].
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 3:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> mat = [[0,0],[1,1],[0,0]]
|
||||
<strong>Output:</strong> [1,2]
|
||||
<strong>Explanation:</strong> The row indexed 1 has the maximum count of ones (2). So the answer is [1,2].
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
<p><strong>Constraints:</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>m == mat.length</code> </li>
|
||||
<li><code>n == mat[i].length</code> </li>
|
||||
<li><code>1 <= m, n <= 100</code> </li>
|
||||
<li><code>mat[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
|
||||
</ul>
|
||||
@ -0,0 +1,41 @@
|
||||
<p>Given the <code>root</code> of a binary tree, replace the value of each node in the tree with the <strong>sum of all its cousins' values</strong>.</p>
|
||||
|
||||
<p>Two nodes of a binary tree are <strong>cousins</strong> if they have the same depth with different parents.</p>
|
||||
|
||||
<p>Return <em>the </em><code>root</code><em> of the modified tree</em>.</p>
|
||||
|
||||
<p><strong>Note</strong> that the depth of a node is the number of edges in the path from the root node to it.</p>
|
||||
|
||||
<p> </p>
|
||||
<p><strong class="example">Example 1:</strong></p>
|
||||
<img alt="" src="https://assets.leetcode.com/uploads/2023/01/11/example11.png" style="width: 571px; height: 151px;" />
|
||||
<pre>
|
||||
<strong>Input:</strong> root = [5,4,9,1,10,null,7]
|
||||
<strong>Output:</strong> [0,0,0,7,7,null,11]
|
||||
<strong>Explanation:</strong> The diagram above shows the initial binary tree and the binary tree after changing the value of each node.
|
||||
- Node with value 5 does not have any cousins so its sum is 0.
|
||||
- Node with value 4 does not have any cousins so its sum is 0.
|
||||
- Node with value 9 does not have any cousins so its sum is 0.
|
||||
- Node with value 1 has a cousin with value 7 so its sum is 7.
|
||||
- Node with value 10 has a cousin with value 7 so its sum is 7.
|
||||
- Node with value 7 has cousins with values 1 and 10 so its sum is 11.
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 2:</strong></p>
|
||||
<img alt="" src="https://assets.leetcode.com/uploads/2023/01/11/diagram33.png" style="width: 481px; height: 91px;" />
|
||||
<pre>
|
||||
<strong>Input:</strong> root = [3,1,2]
|
||||
<strong>Output:</strong> [0,0,0]
|
||||
<strong>Explanation:</strong> The diagram above shows the initial binary tree and the binary tree after changing the value of each node.
|
||||
- Node with value 3 does not have any cousins so its sum is 0.
|
||||
- Node with value 1 does not have any cousins so its sum is 0.
|
||||
- Node with value 2 does not have any cousins so its sum is 0.
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
<p><strong>Constraints:</strong></p>
|
||||
|
||||
<ul>
|
||||
<li>The number of nodes in the tree is in the range <code>[1, 10<sup>5</sup>]</code>.</li>
|
||||
<li><code>1 <= Node.val <= 10<sup>4</sup></code></li>
|
||||
</ul>
|
||||
@ -0,0 +1,38 @@
|
||||
<p>You are given a <strong>0-indexed</strong> array <code>nums</code> consisiting of <strong>positive</strong> integers. You can do the following operation on the array <strong>any</strong> number of times:</p>
|
||||
|
||||
<ul>
|
||||
<li>Select an index <code>i</code> such that <code>0 <= i < n - 1</code> and replace either of <code>nums[i]</code> or <code>nums[i+1]</code> with their gcd value.</li>
|
||||
</ul>
|
||||
|
||||
<p>Return <em>the <strong>minimum</strong> number of operations to make all elements of </em><code>nums</code><em> equal to </em><code>1</code>. If it is impossible, return <code>-1</code>.</p>
|
||||
|
||||
<p>The gcd of two integers is the greatest common divisor of the two integers.</p>
|
||||
|
||||
<p> </p>
|
||||
<p><strong class="example">Example 1:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> nums = [2,6,3,4]
|
||||
<strong>Output:</strong> 4
|
||||
<strong>Explanation:</strong> We can do the following operations:
|
||||
- Choose index i = 2 and replace nums[2] with gcd(3,4) = 1. Now we have nums = [2,6,1,4].
|
||||
- Choose index i = 1 and replace nums[1] with gcd(6,1) = 1. Now we have nums = [2,1,1,4].
|
||||
- Choose index i = 0 and replace nums[0] with gcd(2,1) = 1. Now we have nums = [1,1,1,4].
|
||||
- Choose index i = 2 and replace nums[3] with gcd(1,4) = 1. Now we have nums = [1,1,1,1].
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 2:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> nums = [2,10,6,14]
|
||||
<strong>Output:</strong> -1
|
||||
<strong>Explanation:</strong> It can be shown that it is impossible to make all the elements equal to 1.
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
<p><strong>Constraints:</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>2 <= nums.length <= 50</code></li>
|
||||
<li><code>1 <= nums[i] <= 10<sup>6</sup></code></li>
|
||||
</ul>
|
||||
@ -0,0 +1,35 @@
|
||||
<p>Given a positive integer <code>n</code>, find the sum of all integers in the range <code>[1, n]</code> <strong>inclusive</strong> that are divisible by <code>3</code>, <code>5</code>, or <code>7</code>.</p>
|
||||
|
||||
<p>Return <em>an integer denoting the sum of all numbers in the given range satisfying the constraint.</em></p>
|
||||
|
||||
<p> </p>
|
||||
<p><strong class="example">Example 1:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> n = 7
|
||||
<strong>Output:</strong> 21
|
||||
<strong>Explanation:</strong> Numbers in the range <code>[1, 7]</code> that are divisible by <code>3</code>, <code>5,</code> or <code>7 </code>are <code>3, 5, 6, 7</code>. The sum of these numbers is <code>21</code>.
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 2:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> n = 10
|
||||
<strong>Output:</strong> 40
|
||||
<strong>Explanation:</strong> Numbers in the range <code>[1, 10] that are</code> divisible by <code>3</code>, <code>5,</code> or <code>7</code> are <code>3, 5, 6, 7, 9, 10</code>. The sum of these numbers is 40.
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 3:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> n = 9
|
||||
<strong>Output:</strong> 30
|
||||
<strong>Explanation:</strong> Numbers in the range <code>[1, 9]</code> that are divisible by <code>3</code>, <code>5</code>, or <code>7</code> are <code>3, 5, 6, 7, 9</code>. The sum of these numbers is <code>30</code>.
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
<p><strong>Constraints:</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>1 <= n <= 10<sup>3</sup></code></li>
|
||||
</ul>
|
||||
77
leetcode-cn/problem (English)/函数防抖(English) [debounce].html
Normal file
77
leetcode-cn/problem (English)/函数防抖(English) [debounce].html
Normal file
@ -0,0 +1,77 @@
|
||||
<p>Given a function <code>fn</code> and a time in milliseconds <code>t</code>, return a <strong>debounced</strong> version of that function.</p>
|
||||
|
||||
<p>A <strong>debounced</strong> function is a function whose execution is delayed by <code>t</code> milliseconds and whose execution is cancelled if it is called again within that window of time. The debounced function should also recieve the passed parameters.</p>
|
||||
|
||||
<p>For example, let's say <code>t = 50ms</code>, and the function was called at <code>30ms</code>, <code>60ms</code>, and <code>100ms</code>. The first 2 function calls would be cancelled, and the 3rd function call would be executed at <code>150ms</code>. If instead <code>t = 35ms</code>, The 1st call would be cancelled, the 2nd would be executed at <code>95ms</code>, and the 3rd would be executed at <code>135ms</code>.</p>
|
||||
|
||||
<p><img alt="Debounce Schematic" src="https://assets.leetcode.com/uploads/2023/04/08/screen-shot-2023-04-08-at-11048-pm.png" style="width: 800px; height: 242px;" /></p>
|
||||
|
||||
<p>The above diagram shows how debounce will transform events. Each rectangle represents 100ms and the debounce time is 400ms. Each color represents a different set of inputs.</p>
|
||||
|
||||
<p>Please solve it without using lodash's <code>_.debounce()</code> function.</p>
|
||||
|
||||
<p> </p>
|
||||
<p><strong class="example">Example 1:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong>
|
||||
t = 50
|
||||
calls = [
|
||||
{"t": 50, inputs: [1]},
|
||||
{"t": 75, inputs: [2]}
|
||||
]
|
||||
<strong>Output:</strong> [{"t": 125, inputs: [2]}]
|
||||
<strong>Explanation:</strong>
|
||||
let start = Date.now();
|
||||
function log(...inputs) {
|
||||
console.log([Date.now() - start, inputs ])
|
||||
}
|
||||
const dlog = debounce(log, 50);
|
||||
setTimeout(() => dlog(1), 50);
|
||||
setTimeout(() => dlog(2), 75);
|
||||
|
||||
The 1st call is cancelled by the 2nd call because the 2nd call occurred before 100ms
|
||||
The 2nd call is delayed by 50ms and executed at 125ms. The inputs were (2).
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 2:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong>
|
||||
t = 20
|
||||
calls = [
|
||||
{"t": 50, inputs: [1]},
|
||||
{"t": 100, inputs: [2]}
|
||||
]
|
||||
<strong>Output:</strong> [{"t": 70, inputs: [1]}, {"t": 120, inputs: [2]}]
|
||||
<strong>Explanation:</strong>
|
||||
The 1st call is delayed until 70ms. The inputs were (1).
|
||||
The 2nd call is delayed until 120ms. The inputs were (2).
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 3:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong>
|
||||
t = 150
|
||||
calls = [
|
||||
{"t": 50, inputs: [1, 2]},
|
||||
{"t": 300, inputs: [3, 4]},
|
||||
{"t": 300, inputs: [5, 6]}
|
||||
]
|
||||
<strong>Output:</strong> [{"t": 200, inputs: [1,2]}, {"t": 450, inputs: [5, 6]}]
|
||||
<strong>Explanation:</strong>
|
||||
The 1st call is delayed by 150ms and ran at 200ms. The inputs were (1, 2).
|
||||
The 2nd call is cancelled by the 3rd call
|
||||
The 3rd call is delayed by 150ms and ran at 450ms. The inputs were (5, 6).
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
<p><strong>Constraints:</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>0 <= t <= 1000</code></li>
|
||||
<li><code>1 <= calls.length <= 10</code></li>
|
||||
<li><code>0 <= calls[i].t <= 1000</code></li>
|
||||
<li><code>0 <= calls[i].inputs.length <= 10</code></li>
|
||||
</ul>
|
||||
83
leetcode-cn/problem (English)/分组(English) [group-by].html
Normal file
83
leetcode-cn/problem (English)/分组(English) [group-by].html
Normal file
@ -0,0 +1,83 @@
|
||||
<p>Write code that enhances all arrays such that you can call the <code>array.groupBy(fn)</code> method on any array and it will return a <strong>grouped</strong> version of the array.</p>
|
||||
|
||||
<p>A <strong>grouped</strong> array is an object where each key is the output of <code>fn(arr[i])</code> and each value is an array containing all items in the original array with that key.</p>
|
||||
|
||||
<p>The provided callback <code>fn</code> will accept an item in the array and return a string key.</p>
|
||||
|
||||
<p>The order of each value list should be the order the items appear in the array. Any order of keys is acceptable.</p>
|
||||
|
||||
<p>Please solve it without lodash's <code>_.groupBy</code> function.</p>
|
||||
|
||||
<p> </p>
|
||||
<p><strong class="example">Example 1:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong>
|
||||
array = [
|
||||
{"id":"1"},
|
||||
{"id":"1"},
|
||||
{"id":"2"}
|
||||
],
|
||||
fn = function (item) {
|
||||
return item.id;
|
||||
}
|
||||
<strong>Output:</strong>
|
||||
{
|
||||
"1": [{"id": "1"}, {"id": "1"}],
|
||||
"2": [{"id": "2"}]
|
||||
}
|
||||
<strong>Explanation:</strong>
|
||||
Output is from array.groupBy(fn).
|
||||
The selector function gets the "id" out of each item in the array.
|
||||
There are two objects with an "id" of 1. Both of those objects are put in the first array.
|
||||
There is one object with an "id" of 2. That object is put in the second array.
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 2:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong>
|
||||
array = [
|
||||
[1, 2, 3],
|
||||
[1, 3, 5],
|
||||
[1, 5, 9]
|
||||
]
|
||||
fn = function (list) {
|
||||
return String(list[0]);
|
||||
}
|
||||
<strong>Output:</strong>
|
||||
{
|
||||
"1": [[1, 2, 3], [1, 3, 5], [1, 5, 9]]
|
||||
}
|
||||
<strong>Explanation:</strong>
|
||||
The array can be of any type. In this case, the selector function defines the key as being the first element in the array.
|
||||
All the arrays have 1 as their first element so they are grouped together.
|
||||
{
|
||||
"1": [[1, 2, 3], [1, 3, 5], [1, 5, 9]]
|
||||
}
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 3:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong>
|
||||
array = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
|
||||
fn = function (n) {
|
||||
return String(n > 5);
|
||||
}
|
||||
<strong>Output:</strong>
|
||||
{
|
||||
"true": [6, 7, 8, 9, 10],
|
||||
"false": [1, 2, 3, 4, 5]
|
||||
}
|
||||
<strong>Explanation:</strong>
|
||||
The selector function splits the array by whether each number is greater than 5.
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
<p><strong>Constraints:</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>0 <= array.length <= 10<sup>5</sup></code></li>
|
||||
<li><code>fn returns a string</code></li>
|
||||
</ul>
|
||||
@ -0,0 +1,50 @@
|
||||
<p>Given an array of functions <code>[f<span style="font-size: 10.8333px;">1</span>, f<sub>2</sub>, f<sub>3</sub>, ..., f<sub>n</sub>]</code>, return a new function <code>fn</code> that is the <strong>function composition</strong> of the array of functions.</p>
|
||||
|
||||
<p>The <strong>function composition</strong> of <code>[f(x), g(x), h(x)]</code> is <code>fn(x) = f(g(h(x)))</code>.</p>
|
||||
|
||||
<p>The <strong>function composition</strong> of an empty list of functions is the <strong>identity function</strong> <code>f(x) = x</code>.</p>
|
||||
|
||||
<p>You may assume each function in the array accepts one integer as input and returns one integer as output.</p>
|
||||
|
||||
<p> </p>
|
||||
<p><strong class="example">Example 1:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> functions = [x => x + 1, x => x * x, x => 2 * x], x = 4
|
||||
<strong>Output:</strong> 65
|
||||
<strong>Explanation:</strong>
|
||||
Evaluating from right to left ...
|
||||
Starting with x = 4.
|
||||
2 * (4) = 8
|
||||
(8) * (8) = 64
|
||||
(64) + 1 = 65
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 2:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> functions = [x => 10 * x, x => 10 * x, x => 10 * x], x = 1
|
||||
<strong>Output:</strong> 1000
|
||||
<strong>Explanation:</strong>
|
||||
Evaluating from right to left ...
|
||||
10 * (1) = 10
|
||||
10 * (10) = 100
|
||||
10 * (100) = 1000
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 3:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> functions = [], x = 42
|
||||
<strong>Output:</strong> 42
|
||||
<strong>Explanation:</strong>
|
||||
The composition of zero functions is the identity function</pre>
|
||||
|
||||
<p> </p>
|
||||
<p><strong>Constraints:</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code><font face="monospace">-1000 <= x <= 1000</font></code></li>
|
||||
<li><code><font face="monospace">0 <= functions.length <= 1000</font></code></li>
|
||||
<li><font face="monospace"><code>all functions accept and return a single integer</code></font></li>
|
||||
</ul>
|
||||
@ -0,0 +1,48 @@
|
||||
<p>Given two objects <code>o1</code> and <code>o2</code>, check if they are <strong>deeply equal</strong>.</p>
|
||||
|
||||
<p>For two objects to be <strong>deeply equal</strong>, they must contain the same keys, and the associated values must also be <strong>deeply equal</strong>. Two objects are also considered <strong>deeply equal</strong> if they pass the <code>===</code> equality check.</p>
|
||||
|
||||
<p>You may assume both objects are the output of <code>JSON.parse</code>. In other words, they are valid JSON.</p>
|
||||
|
||||
<p>Please solve it without using lodash's <code>_.isEqual()</code> function.</p>
|
||||
|
||||
<p> </p>
|
||||
<p><strong class="example">Example 1:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> o1 = {"x":1,"y":2}, o2 = {"x":1,"y":2}
|
||||
<strong>Output:</strong> true
|
||||
<strong>Explanation:</strong> The keys and values match exactly.
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 2:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> o1 = {"y":2,"x":1}, o2 = {"x":1,"y":2}
|
||||
<strong>Output:</strong> true
|
||||
<strong>Explanation:</strong> Although the keys are in a different order, they still match exactly.
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 3:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> o1 = {"x":null,"L":[1,2,3]}, o2 = {"x":null,"L":["1","2","3"]}
|
||||
<strong>Output:</strong> false
|
||||
<strong>Explanation:</strong> The array of numbers is different from the array of strings.
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 4:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> o1 = true, o2 = false
|
||||
<strong>Output:</strong> false
|
||||
<strong>Explanation:</strong> true !== false</pre>
|
||||
|
||||
<p> </p>
|
||||
<p><strong>Constraints:</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>1 <= JSON.stringify(o1).length <= 10<sup>5</sup></code></li>
|
||||
<li><code>1 <= JSON.stringify(o2).length <= 10<sup>5</sup></code></li>
|
||||
<li><code>maxNestingDepth <= 1000</code></li>
|
||||
</ul>
|
||||
@ -0,0 +1,49 @@
|
||||
<p>Given an object, return a valid JSON string of that object. You may assume the object only inludes strings, integers, arrays, objects, booleans, and null. The returned string should not include extra spaces. The order of keys should be the same as the order returned by <code>Object.keys()</code>.</p>
|
||||
|
||||
<p>Please solve it without using the built-in <code>JSON.stringify</code> method.</p>
|
||||
|
||||
<p> </p>
|
||||
<p><strong class="example">Example 1:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> object = {"y":1,"x":2}
|
||||
<strong>Output:</strong> {"y":1,"x":2}
|
||||
<strong>Explanation:</strong>
|
||||
Return the JSON representation.
|
||||
Note that the order of keys should be the same as the order returned by Object.keys().</pre>
|
||||
|
||||
<p><strong class="example">Example 2:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> object = {"a":"str","b":-12,"c":true,"d":null}
|
||||
<strong>Output:</strong> {"a":"str","b":-12,"c":true,"d":null}
|
||||
<strong>Explanation:</strong>
|
||||
The primitives of JSON are strings, numbers, booleans, and null.
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 3:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> object = {"key":{"a":1,"b":[{},null,"Hello"]}}
|
||||
<strong>Output:</strong> {"key":{"a":1,"b":[{},null,"Hello"]}}
|
||||
<strong>Explanation:</strong>
|
||||
Objects and arrays can include other objects and arrays.
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 4:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> object = true
|
||||
<strong>Output:</strong> true
|
||||
<strong>Explanation:</strong>
|
||||
Primitive types are valid inputs.</pre>
|
||||
|
||||
<p> </p>
|
||||
<p><strong>Constraints:</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>object includes strings, integers, booleans, arrays, objects, and null</code></li>
|
||||
<li><code>1 <= JSON.stringify(object).length <= 10<sup>5</sup></code></li>
|
||||
<li><code>maxNestingLevel <= 1000</code></li>
|
||||
<li><code>all strings will only contain alphanumeric characters</code></li>
|
||||
</ul>
|
||||
@ -0,0 +1,36 @@
|
||||
<p>Given a <strong>multi-dimensional array</strong> of integers, return a generator object which yields integers in the same order as <strong>inorder traversal</strong>.</p>
|
||||
|
||||
<p>A <strong>multi-dimensional array</strong> is a recursive data structure that contains both integers and other <strong>multi-dimensional arrays</strong>.</p>
|
||||
|
||||
<p><strong>inorder traversal</strong> iterates over each array from left to right, yielding any integers it encounters or applying <strong>inorder traversal</strong> to any arrays it encounters.</p>
|
||||
|
||||
<p> </p>
|
||||
<p><strong class="example">Example 1:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> arr = [[[6]],[1,3],[]]
|
||||
<strong>Output:</strong> [6,1,3]
|
||||
<strong>Explanation:</strong>
|
||||
const generator = inorderTraversal(arr);
|
||||
generator.next().value; // 6
|
||||
generator.next().value; // 1
|
||||
generator.next().value; // 3
|
||||
generator.next().done; // true
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 2:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> arr = []
|
||||
<strong>Output:</strong> []
|
||||
<strong>Explanation:</strong> There are no integers so the generator doesn't yield anything.
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
<p><strong>Constraints:</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>0 <= arr.flat().length <= 10<sup>5</sup></code></li>
|
||||
<li><code>0 <= arr.flat()[i] <= 10<sup>5</sup></code></li>
|
||||
<li><code>maxNestingDepth <= 10<sup>5</sup></code></li>
|
||||
</ul>
|
||||
@ -0,0 +1,49 @@
|
||||
<p>You are given two <strong>0-indexed</strong> integer arrays <code>nums</code> and <code>divisors</code>.</p>
|
||||
|
||||
<p>The <strong>divisibility score</strong> of <code>divisors[i]</code> is the number of indices <code>j</code> such that <code>nums[j]</code> is divisible by <code>divisors[i]</code>.</p>
|
||||
|
||||
<p>Return <em>the integer</em> <code>divisors[i]</code> <em>with the maximum divisibility score</em>. If there is more than one integer with the maximum score, return the minimum of them.</p>
|
||||
|
||||
<p> </p>
|
||||
<p><strong class="example">Example 1:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> nums = [4,7,9,3,9], divisors = [5,2,3]
|
||||
<strong>Output:</strong> 3
|
||||
<strong>Explanation:</strong> The divisibility score for every element in divisors is:
|
||||
The divisibility score of divisors[0] is 0 since no number in nums is divisible by 5.
|
||||
The divisibility score of divisors[1] is 1 since nums[0] is divisible by 2.
|
||||
The divisibility score of divisors[2] is 3 since nums[2], nums[3], and nums[4] are divisible by 3.
|
||||
Since divisors[2] has the maximum divisibility score, we return it.
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 2:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> nums = [20,14,21,10], divisors = [5,7,5]
|
||||
<strong>Output:</strong> 5
|
||||
<strong>Explanation:</strong> The divisibility score for every element in divisors is:
|
||||
The divisibility score of divisors[0] is 2 since nums[0] and nums[3] are divisible by 5.
|
||||
The divisibility score of divisors[1] is 2 since nums[1] and nums[2] are divisible by 7.
|
||||
The divisibility score of divisors[2] is 2 since nums[0] and nums[3] are divisible by 5.
|
||||
Since divisors[0], divisors[1], and divisors[2] all have the maximum divisibility score, we return the minimum of them (i.e., divisors[2]).
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 3:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> nums = [12], divisors = [10,16]
|
||||
<strong>Output:</strong> 10
|
||||
<strong>Explanation:</strong> The divisibility score for every element in divisors is:
|
||||
The divisibility score of divisors[0] is 0 since no number in nums is divisible by 10.
|
||||
The divisibility score of divisors[1] is 0 since no number in nums is divisible by 16.
|
||||
Since divisors[0] and divisors[1] both have the maximum divisibility score, we return the minimum of them (i.e., divisors[0]).
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
<p><strong>Constraints:</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>1 <= nums.length, divisors.length <= 1000</code></li>
|
||||
<li><code>1 <= nums[i], divisors[i] <= 10<sup>9</sup></code></li>
|
||||
</ul>
|
||||
@ -0,0 +1,25 @@
|
||||
Write code that enhances all arrays such that you can call the <code>array.last()</code> method on any array and it will return the last element. If there are no elements in the array, it should return <code>-1</code>.
|
||||
<p> </p>
|
||||
<p><strong class="example">Example 1:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> nums = [1,2,3]
|
||||
<strong>Output:</strong> 3
|
||||
<strong>Explanation:</strong> Calling nums.last() should return the last element: 3.
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 2:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> nums = []
|
||||
<strong>Output:</strong> -1
|
||||
<strong>Explanation:</strong> Because there are no elements, return -1.
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
<p><strong>Constraints:</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>0 <= arr.length <= 1000</code></li>
|
||||
<li><code>0 <= arr[i] <= 1000</code></li>
|
||||
</ul>
|
||||
@ -0,0 +1,62 @@
|
||||
<p>Given an integer array <code>nums</code>, a reducer function <code>fn</code>, and an intial value <code>init</code>, return a <strong>reduced</strong> array.</p>
|
||||
|
||||
<p>A <strong>reduced</strong> array is created by applying the following operation: <code>val = fn(init, nums[0])</code>, <code>val = fn(val, nums[1])</code>, <code>val = fn(val, arr[2])</code>, <code>...</code> until every element in the array has been processed. The final value of <code>val</code> is returned.</p>
|
||||
|
||||
<p>If the length of the array is 0, it should return <code>init</code>.</p>
|
||||
|
||||
<p>Please solve it without using the built-in <code>Array.reduce</code> method.</p>
|
||||
|
||||
<p> </p>
|
||||
<p><strong class="example">Example 1:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong>
|
||||
nums = [1,2,3,4]
|
||||
fn = function sum(accum, curr) { return accum + curr; }
|
||||
init = 0
|
||||
<strong>Output:</strong> 10
|
||||
<strong>Explanation:</strong>
|
||||
initially, the value is init=0.
|
||||
(0) + nums[0] = 1
|
||||
(1) + nums[1] = 3
|
||||
(3) + nums[2] = 6
|
||||
(6) + nums[3] = 10
|
||||
The final answer is 10.
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 2:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong>
|
||||
nums = [1,2,3,4]
|
||||
fn = function sum(accum, curr) { return accum + curr * curr; }
|
||||
init = 100
|
||||
<strong>Output:</strong> 130
|
||||
<strong>Explanation:</strong>
|
||||
initially, the value is init=100.
|
||||
(100) + nums[0]^2 = 101
|
||||
(101) + nums[1]^2 = 105
|
||||
(105) + nums[2]^2 = 114
|
||||
(114) + nums[3]^2 = 130
|
||||
The final answer is 130.
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 3:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong>
|
||||
nums = []
|
||||
fn = function sum(accum, curr) { return 0; }
|
||||
init = 25
|
||||
<strong>Output:</strong> 25
|
||||
<strong>Explanation:</strong> For empty arrays, the answer is always init.
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
<p><strong>Constraints:</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>0 <= nums.length <= 1000</code></li>
|
||||
<li><code>0 <= nums[i] <= 1000</code></li>
|
||||
<li><code>0 <= init <= 1000</code></li>
|
||||
</ul>
|
||||
@ -0,0 +1,50 @@
|
||||
<p>There exists an undirected and unrooted tree with <code>n</code> nodes indexed from <code>0</code> to <code>n - 1</code>. You are given the integer <code>n</code> and a 2D integer array <code>edges</code> of length <code>n - 1</code>, where <code>edges[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> indicates that there is an edge between nodes <code>a<sub>i</sub></code> and <code>b<sub>i</sub></code> in the tree.</p>
|
||||
|
||||
<p>Each node has an associated price. You are given an integer array <code>price</code>, where <code>price[i]</code> is the price of the <code>i<sup>th</sup></code> node.</p>
|
||||
|
||||
<p>The <strong>price sum</strong> of a given path is the sum of the prices of all nodes lying on that path.</p>
|
||||
|
||||
<p>Additionally, you are given a 2D integer array <code>trips</code>, where <code>trips[i] = [start<sub>i</sub>, end<sub>i</sub>]</code> indicates that you start the <code>i<sup>th</sup></code> trip from the node <code>start<sub>i</sub></code> and travel to the node <code>end<sub>i</sub></code> by any path you like.</p>
|
||||
|
||||
<p>Before performing your first trip, you can choose some <strong>non-adjacent</strong> nodes and halve the prices.</p>
|
||||
|
||||
<p>Return <em>the minimum total price sum to perform all the given trips</em>.</p>
|
||||
|
||||
<p> </p>
|
||||
<p><strong class="example">Example 1:</strong></p>
|
||||
<img alt="" src="https://assets.leetcode.com/uploads/2023/03/16/diagram2.png" style="width: 541px; height: 181px;" />
|
||||
<pre>
|
||||
<strong>Input:</strong> n = 4, edges = [[0,1],[1,2],[1,3]], price = [2,2,10,6], trips = [[0,3],[2,1],[2,3]]
|
||||
<strong>Output:</strong> 23
|
||||
<strong>Explanation:</strong> The diagram above denotes the tree after rooting it at node 2. The first part shows the initial tree and the second part shows the tree after choosing nodes 0, 2, and 3, and making their price half.
|
||||
For the 1<sup>st</sup> trip, we choose path [0,1,3]. The price sum of that path is 1 + 2 + 3 = 6.
|
||||
For the 2<sup>nd</sup> trip, we choose path [2,1]. The price sum of that path is 2 + 5 = 7.
|
||||
For the 3<sup>rd</sup> trip, we choose path [2,1,3]. The price sum of that path is 5 + 2 + 3 = 10.
|
||||
The total price sum of all trips is 6 + 7 + 10 = 23.
|
||||
It can be proven, that 23 is the minimum answer that we can achieve.
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 2:</strong></p>
|
||||
<img alt="" src="https://assets.leetcode.com/uploads/2023/03/16/diagram3.png" style="width: 456px; height: 111px;" />
|
||||
<pre>
|
||||
<strong>Input:</strong> n = 2, edges = [[0,1]], price = [2,2], trips = [[0,0]]
|
||||
<strong>Output:</strong> 1
|
||||
<strong>Explanation:</strong> The diagram above denotes the tree after rooting it at node 0. The first part shows the initial tree and the second part shows the tree after choosing node 0, and making its price half.
|
||||
For the 1<sup>st</sup> trip, we choose path [0]. The price sum of that path is 1.
|
||||
The total price sum of all trips is 1. It can be proven, that 1 is the minimum answer that we can achieve.
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
<p><strong>Constraints:</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>1 <= n <= 50</code></li>
|
||||
<li><code>edges.length == n - 1</code></li>
|
||||
<li><code>0 <= a<sub>i</sub>, b<sub>i</sub> <= n - 1</code></li>
|
||||
<li><code>edges</code> represents a valid tree.</li>
|
||||
<li><code>price.length == n</code></li>
|
||||
<li><code>price[i]</code> is an even integer.</li>
|
||||
<li><code>1 <= price[i] <= 1000</code></li>
|
||||
<li><code>1 <= trips.length <= 100</code></li>
|
||||
<li><code>0 <= start<sub>i</sub>, end<sub>i</sub> <= n - 1</code></li>
|
||||
</ul>
|
||||
@ -0,0 +1,71 @@
|
||||
<p>Given an asyncronous function <code>fn</code> and a time <code>t</code> in milliseconds, return a new <strong>time limited</strong> version of the input function.</p>
|
||||
|
||||
<p>A <strong>time limited</strong> function is a function that is identical to the original unless it takes longer than <code>t</code> milliseconds to fullfill. In that case, it will reject with <code>"Time Limit Exceeded"</code>. Note that it should reject with a string, not an <code>Error</code>.</p>
|
||||
|
||||
<p> </p>
|
||||
<p><strong class="example">Example 1:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong>
|
||||
fn = async (n) => {
|
||||
await new Promise(res => setTimeout(res, 100));
|
||||
return n * n;
|
||||
}
|
||||
inputs = [5]
|
||||
t = 50
|
||||
<strong>Output:</strong> {"rejected":"Time Limit Exceeded","time":50}
|
||||
<strong>Explanation:</strong>
|
||||
The provided function is set to resolve after 100ms. However, the time limit is set to 50ms. It rejects at t=50ms because the time limit was reached.
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 2:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong>
|
||||
fn = async (n) => {
|
||||
await new Promise(res => setTimeout(res, 100));
|
||||
return n * n;
|
||||
}
|
||||
inputs = [5]
|
||||
t = 150
|
||||
<strong>Output:</strong> {"resolved":25,"time":100}
|
||||
<strong>Explanation:</strong>
|
||||
The function resolved 5 * 5 = 25 at t=100ms. The time limit is never reached.
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 3:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong>
|
||||
fn = async (a, b) => {
|
||||
await new Promise(res => setTimeout(res, 120));
|
||||
return a + b;
|
||||
}
|
||||
inputs = [5,10]
|
||||
t = 150
|
||||
<strong>Output:</strong> {"resolved":15,"time":120}
|
||||
<strong>Explanation:</strong>
|
||||
The function resolved 5 + 10 = 15 at t=120ms. The time limit is never reached.
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 4:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong>
|
||||
fn = async () => {
|
||||
throw "Error";
|
||||
}
|
||||
inputs = []
|
||||
t = 1000
|
||||
<strong>Output:</strong> {"rejected":"Error","time":0}
|
||||
<strong>Explanation:</strong>
|
||||
The function immediately throws an error.</pre>
|
||||
|
||||
<p> </p>
|
||||
<p><strong>Constraints:</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>0 <= inputs.length <= 10</code></li>
|
||||
<li><code>0 <= t <= 1000</code></li>
|
||||
<li><code>fn returns a promise</code></li>
|
||||
</ul>
|
||||
@ -0,0 +1,56 @@
|
||||
<p>Write a class that allows getting and setting key-value pairs, however a <strong>time until expiration</strong> is associated with each key.</p>
|
||||
|
||||
<p>The class has three public methods:</p>
|
||||
|
||||
<p><code>set(key, value, duration)</code>: accepts an integer <code>key</code>, an integer <code>value</code>, and a <code>duration</code> in milliseconds. Once the <code>duration</code> has elapsed, the key should be inaccessible. The method should return <code>true</code> if the same un-expired key already exists and <code>false</code> otherwise. Both the value and duration should be overwritten if the key already exists.</p>
|
||||
|
||||
<p><code>get(key)</code>: if an un-expired key exists, it should return the associated value. Otherwise it should return <code>-1</code>.</p>
|
||||
|
||||
<p><code>count()</code>: returns the count of un-expired keys.</p>
|
||||
|
||||
<p> </p>
|
||||
<p><strong class="example">Example 1:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong>
|
||||
["TimeLimitedCache", "set", "get", "count", "get"]
|
||||
[[], [1, 42, 100], [1], [], [1]]
|
||||
[0, 0, 50, 50, 150]
|
||||
<strong>Output:</strong> [null, false, 42, 1, -1]
|
||||
<strong>Explanation:</strong>
|
||||
At t=0, the cache is constructed.
|
||||
At t=0, a key-value pair (1: 42) is added with a time limit of 100ms. The value doesn't exist so false is returned.
|
||||
At t=50, key=1 is requested and the value of 42 is returned.
|
||||
At t=50, count() is called and there is one active key in the cache.
|
||||
At t=100, key=1 expires.
|
||||
At t=150, get(1) is called but -1 is returned because the cache is empty.
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 2:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong>
|
||||
["TimeLimitedCache", "set", "set", "get", "get", "get", "count"]
|
||||
[[], [1, 42, 50], [1, 50, 100], [1], [1], [1], []]
|
||||
[0, 0, 40, 50, 120, 200, 250]
|
||||
<strong>Output:</strong> [null, false, true, 50, 50, -1]
|
||||
<strong>Explanation:</strong>
|
||||
At t=0, the cache is constructed.
|
||||
At t=0, a key-value pair (1: 42) is added with a time limit of 50ms. The value doesn't exist so false is returned.
|
||||
At t=40, a key-value pair (1: 50) is added with a time limit of 100ms. A non-expired value already existed so true is returned and the old value was overwritten.
|
||||
At t=50, get(1) is called which returned 50.
|
||||
At t=120, get(1) is called which returned 50.
|
||||
At t=140, key=1 expires.
|
||||
At t=200, get(1) is called but the cache is empty so -1 is returned.
|
||||
At t=250, count() returns 0 because the cache is empty.
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
<p><strong>Constraints:</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>0 <= key <= 10<sup>9</sup></code></li>
|
||||
<li><code>0 <= value <= 10<sup>9</sup></code></li>
|
||||
<li><code>0 <= duration <= 1000</code></li>
|
||||
<li><code>total method calls will not exceed 100</code></li>
|
||||
</ul>
|
||||
@ -0,0 +1,36 @@
|
||||
<p>Given a string <code>word</code> to which you can insert letters "a", "b" or "c" anywhere and any number of times, return <em>the minimum number of letters that must be inserted so that <code>word</code> becomes <strong>valid</strong>.</em></p>
|
||||
|
||||
<p>A string is called <strong>valid </strong>if it can be formed by concatenating the string "abc" several times.</p>
|
||||
|
||||
<p> </p>
|
||||
<p><strong class="example">Example 1:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> word = "b"
|
||||
<strong>Output:</strong> 2
|
||||
<strong>Explanation:</strong> Insert the letter "a" right before "b", and the letter "c" right next to "a" to obtain the valid string "<strong>a</strong>b<strong>c</strong>".
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 2:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> word = "aaa"
|
||||
<strong>Output:</strong> 6
|
||||
<strong>Explanation:</strong> Insert letters "b" and "c" next to each "a" to obtain the valid string "a<strong>bc</strong>a<strong>bc</strong>a<strong>bc</strong>".
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 3:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> word = "abc"
|
||||
<strong>Output:</strong> 0
|
||||
<strong>Explanation:</strong> word is already valid. No modifications are needed.
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
<p><strong>Constraints:</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>1 <= word.length <= 50</code></li>
|
||||
<li><code>word</code> consists of letters "a", "b" and "c" only. </li>
|
||||
</ul>
|
||||
@ -0,0 +1,39 @@
|
||||
<p>You are given a <strong>0-indexed</strong> <code>m x n</code> integer matrix <code>grid</code>. The width of a column is the maximum <strong>length </strong>of its integers.</p>
|
||||
|
||||
<ul>
|
||||
<li>For example, if <code>grid = [[-10], [3], [12]]</code>, the width of the only column is <code>3</code> since <code>-10</code> is of length <code>3</code>.</li>
|
||||
</ul>
|
||||
|
||||
<p>Return <em>an integer array</em> <code>ans</code> <em>of size</em> <code>n</code> <em>where</em> <code>ans[i]</code> <em>is the width of the</em> <code>i<sup>th</sup></code> <em>column</em>.</p>
|
||||
|
||||
<p>The <strong>length</strong> of an integer <code>x</code> with <code>len</code> digits is equal to <code>len</code> if <code>x</code> is non-negative, and <code>len + 1</code> otherwise.</p>
|
||||
|
||||
<p> </p>
|
||||
<p><strong class="example">Example 1:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> grid = [[1],[22],[333]]
|
||||
<strong>Output:</strong> [3]
|
||||
<strong>Explanation:</strong> In the 0<sup>th</sup> column, 333 is of length 3.
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 2:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> grid = [[-15,1,3],[15,7,12],[5,6,-2]]
|
||||
<strong>Output:</strong> [3,1,2]
|
||||
<strong>Explanation:</strong>
|
||||
In the 0<sup>th</sup> column, only -15 is of length 3.
|
||||
In the 1<sup>st</sup> column, all integers are of length 1.
|
||||
In the 2<sup>nd</sup> column, both 12 and -2 are of length 2.
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
<p><strong>Constraints:</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>m == grid.length</code></li>
|
||||
<li><code>n == grid[i].length</code></li>
|
||||
<li><code>1 <= m, n <= 100 </code></li>
|
||||
<li><code>-10<sup>9</sup> <= grid[r][c] <= 10<sup>9</sup></code></li>
|
||||
</ul>
|
||||
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Reference in New Issue
Block a user