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https://gitee.com/coder-xiaomo/leetcode-problemset
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"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 <code>Array.reduce</code> 前提下解决这个问题。</p>\n\n<p> </p>\n\n<p><strong class=\"example\">示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>\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> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>0 <= nums.length <= 1000</code></li>\n\t<li><code>0 <= nums[i] <= 1000</code></li>\n\t<li><code>0 <= init <= 1000</code></li>\n</ul>\n",
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"Declare a variable \"res\" and set it it equal to the initial value.",
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"Loop over each value in the array and set \"res\" = fn(res, arr[i])."
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