leetcode-problemset/leetcode-cn/originData/array-reduce-transformation.json

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            "questionId": "2761",
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            "categoryTitle": "JavaScript",
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            "title": "Array Reduce Transformation",
            "titleSlug": "array-reduce-transformation",
            "content": "<p>Given an integer array <code>nums</code>, a reducer function <code>fn</code>, and an initial value <code>init</code>, return the final result obtained by executing the <code>fn</code> function on each element of the array, sequentially, passing in the return value from the calculation on the preceding element.</p>\n\n<p>This result is achieved through the following operations: <code>val = fn(init, nums[0]), val = fn(val, nums[1]), val = fn(val, nums[2]), ...</code> until every element in the array has been processed. The ultimate value of <code>val</code> is then returned.</p>\n\n<p>If the length of the array is 0, the function should return <code>init</code>.</p>\n\n<p>Please solve it without using the built-in <code>Array.reduce</code> method.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> \nnums = [1,2,3,4]\nfn = function sum(accum, curr) { return accum + curr; }\ninit = 0\n<strong>Output:</strong> 10\n<strong>Explanation:</strong>\ninitially, the value is init=0.\n(0) + nums[0] = 1\n(1) + nums[1] = 3\n(3) + nums[2] = 6\n(6) + nums[3] = 10\nThe final answer is 10.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> \nnums = [1,2,3,4]\nfn = function sum(accum, curr) { return accum + curr * curr; }\ninit = 100\n<strong>Output:</strong> 130\n<strong>Explanation:</strong>\ninitially, the value is init=100.\n(100) + nums[0] * nums[0] = 101\n(101) + nums[1] * nums[1] = 105\n(105) + nums[2] * nums[2] = 114\n(114) + nums[3] * nums[3] = 130\nThe final answer is 130.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> \nnums = []\nfn = function sum(accum, curr) { return 0; }\ninit = 25\n<strong>Output:</strong> 25\n<strong>Explanation:</strong> For empty arrays, the answer is always init.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>0 &lt;= nums.length &lt;= 1000</code></li>\n\t<li><code>0 &lt;= nums[i] &lt;= 1000</code></li>\n\t<li><code>0 &lt;= init &lt;= 1000</code></li>\n</ul>\n",
            "translatedTitle": "数组归约运算",
            "translatedContent": "<p>给定一个整数数组 <code>nums</code>、一个 reducer 函数 <code>fn</code> 和一个初始值 <code>init</code>，返回通过依次对数组的每个元素执行 <code>fn</code> 函数得到的最终结果。</p>\n\n<p>通过以下操作实现这个结果：<code>val = fn(init, nums[0])，val = fn(val, nums[1])，val = fn(val, nums[2])，...</code> 直到处理数组中的每个元素。然后返回 <code>val</code> 的最终值。</p>\n\n<p>如果数组的长度为 0，则函数应返回 <code>init</code>。</p>\n\n<p>请你在不使用内置数组方法的&nbsp;<code>Array.reduce</code>&nbsp;前提下解决这个问题。</p>\n\n<p>&nbsp;</p>\n\n<p><strong class=\"example\">示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>\nnums = [1,2,3,4]\nfn = function sum(accum, curr) { return accum + curr; }\ninit = 0\n<strong>输出：</strong>10\n<strong>解释：</strong>\n初始值为 init=0 。\n(0) + nums[0] = 1\n(1) + nums[1] = 3\n(3) + nums[2] = 6\n(6) + nums[3] = 10\nVal 最终值为 10。\n</pre>\n\n<p><strong class=\"example\">示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong> \nnums = [1,2,3,4]\nfn = function sum(accum, curr) { return accum + curr * curr; }\ninit = 100\n<strong>输出：</strong>130\n<strong>解释：</strong>\n初始值为 init=100 。\n(100) + nums[0]^2 = 101\n(101) + nums[1]^2 = 105\n(105) + nums[2]^2 = 114\n(114) + nums[3]^2 = 130\nVal 最终值为 130。\n</pre>\n\n<p><strong class=\"example\">示例3:</strong></p>\n\n<pre>\n<strong>输入：</strong> \nnums = []\nfn = function sum(accum, curr) { return 0; }\ninit = 25\n<strong>输出：</strong>25\n<b>解释：</b>这是一个空数组，所以返回 init 。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>0 &lt;= nums.length &lt;= 1000</code></li>\n\t<li><code>0 &lt;= nums[i] &lt;= 1000</code></li>\n\t<li><code>0 &lt;= init &lt;= 1000</code></li>\n</ul>\n",
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                    "code": "/**\n * @param {number[]} nums\n * @param {Function} fn\n * @param {number} init\n * @return {number}\n */\nvar reduce = function(nums, fn, init) {\n    \n};",
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