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docs: 第六题答案
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```js
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/**
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* @param input
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* @param size
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* @returns {Array}
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*/
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_.chunk(['a', 'b', 'c', 'd'], 2)
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// => [['a', 'b'], ['c', 'd']]
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_.chunk(['a', 'b', 'c', 'd'], 3)
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// => [['a', 'b', 'c'], ['d']]
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_.chunk(['a', 'b', 'c', 'd'], 5)
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// => [['a', 'b', 'c', 'd']]
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_.chunk(['a', 'b', 'c', 'd'], 0)
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// => []
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```
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这道题目其实就是让大家实现一个 lodash 里的函数,这边我们需要注意的一个点是不能更改原数组,虽然题目没有提到,但是我们得想到这个。
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笔者这里推荐几个答案,分别是不同的写法,难度从低到高吧。
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首先是遍历的写法:
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[答案链接](https://github.com/KieSun/fucking-frontend/issues/8#issuecomment-799053150)
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```js
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function chunk(arr, num) {
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if (num === 0) return [];
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if (Array.isArray(arr) && typeof num === "number") {
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let result = [];
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let i = 0;
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while (i < arr.length) {
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result.push(arr.slice(i, i + num));
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i += num;
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}
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return result;
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} else {
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console.log("params type error");
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}
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}
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```
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我们也可以使用 `reduce` 来解题:
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[答案链接](https://github.com/KieSun/fucking-frontend/issues/8#issuecomment-799021398)
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```js
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const chunk = (arr, len) => arr.reduce((pre, cur, index) => {
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if (index % len === 0) {
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pre.push([cur])
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return pre
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}
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const temp = pre[pre.length - 1]
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temp && temp.push(cur)
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return pre
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}, [])
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```
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甚至直接优化到一行:
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[答案链接](https://github.com/KieSun/fucking-frontend/issues/8#issuecomment-799344324)
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```js
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function chunk(arr, size){
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return Array.from({length: (size = Number.parseInt(size)) ? Math.ceil(arr.length/size) : 0})
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.map((a,i) => arr.slice(i*size, (i+1)*size))
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}
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```
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