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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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62
leetcode-cn/originData/array-prototype-last.json
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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",
|
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"translatedTitle": "数组原型对象的最后一个元素",
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||||
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62
leetcode-cn/originData/array-reduce-transformation.json
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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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||||
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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
File diff suppressed because one or more lines are too long
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||||
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||||
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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
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||||
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||||
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||||
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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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63
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
Normal file
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
Normal file
63
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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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||||
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||||
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64
leetcode-cn/originData/memoize-ii.json
Normal file
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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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"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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||||
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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
File diff suppressed because one or more lines are too long
176
leetcode-cn/originData/row-with-maximum-ones.json
Normal file
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
@@ -0,0 +1,63 @@
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||||
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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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||||
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||||
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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
@@ -0,0 +1,62 @@
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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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||||
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163
leetcode-cn/originData/sum-multiples.json
Normal file
163
leetcode-cn/originData/sum-multiples.json
Normal file
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161
leetcode-cn/originData/xepqZ5.json
Normal file
161
leetcode-cn/originData/xepqZ5.json
Normal file
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Reference in New Issue
Block a user