mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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62 lines
11 KiB
JSON
62 lines
11 KiB
JSON
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"question": {
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"questionId": "2740",
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"questionFrontendId": "2632",
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"categoryTitle": "JavaScript",
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"boundTopicId": 2222271,
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"title": "Curry",
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"titleSlug": "curry",
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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",
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"translatedTitle": "柯里化",
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"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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"difficulty": "Medium",
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"code": "/**\n * @param {Function} fn\n * @return {Function}\n */\nvar curry = function(fn) {\n return function curried() {\n\n };\n};\n\n/**\n * function sum(a, b) { return a + b; }\n * const csum = curry(sum);\n * csum(1)(2) // 3\n */",
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"code": "function curry(fn: Function): Function {\n return function curried() {\n\n };\n};\n\n/**\n * function sum(a, b) { return a + b; }\n * const csum = curry(sum);\n * csum(1)(2) // 3\n */",
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"hints": [
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"You can access the count of parameters expected to passed into a function with \"fn.length\".",
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"You can use recursion. If the length of params passed is equal to fn.length, you are done. Just pass those params to fn. Otherwise return a function that is includes the previous passed params plus the new params. The new function should contain a recursive call to curry()."
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"sampleTestCase": "function sum(a, b, c) { return a + b + c; }\n[[1],[2],[3]]",
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"exampleTestcases": "function sum(a, b, c) { return a + b + c; }\n[[1],[2],[3]]\nfunction sum(a, b, c) { return a + b + c; }\n[[1,2],[3]]\nfunction sum(a, b, c) { return a + b + c; }\n[[],[],[1,2,3]]\nfunction life() { return 42; }\n[[]]",
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