220 lines
10 KiB
Markdown
220 lines
10 KiB
Markdown
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---
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title: Chunky Monkey
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localeTitle: 矮胖的猴子
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---
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![](//discourse-user-assets.s3.amazonaws.com/original/2X/a/aadd6bead83ab7d79a795c326f005a89e6ad81f5.png)
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![:triangular_flag_on_post:](https://forum.freecodecamp.com/images/emoji/emoji_one/triangular_flag_on_post.png?v=3 ":triangular_flag_on_post:")如果卡住,请记得使用**`Read-Search-Ask`** 。尝试配对程序![:busts_in_silhouette:](https://forum.freecodecamp.com/images/emoji/emoji_one/busts_in_silhouette.png?v=3 ":busts_in_silhouette:")并编写自己的代码![:pencil:](https://forum.freecodecamp.com/images/emoji/emoji_one/pencil.png?v=3 ":铅笔:")
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### ![:checkered_flag:](https://forum.freecodecamp.com/images/emoji/emoji_one/checkered_flag.png?v=3 ":checkered_flag:")问题说明:
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我们对此算法的目标是将`arr` (第一个参数)拆分为较小的数组块,其长度由`size` (第二个参数)提供。我们的代码需要传递4个绿色检查(目标)才能完成此算法:
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1. `(['a', 'b', 'c', 'd'], 2)`预期为`[['a', 'b'], ['c', 'd']]`
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2. `([0, 1, 2, 3, 4, 5], 3)`预计`[[0, 1, 2], [3, 4, 5]]`
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3. `([0, 1, 2, 3, 4, 5], 2)`预计`[[0, 1], [2, 3], [4, 5]]`
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4. `([0, 1, 2, 3, 4, 5], 4)`预计`[[0, 1, 2, 3], [4, 5]]`
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#### 相关链接
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* [的Array.push()](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/push)
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* [Array.slice()](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/slice)
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## ![:speech_balloon:](https://forum.freecodecamp.com/images/emoji/emoji_one/speech_balloon.png?v=3 ":speech_balloon:")提示:1
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上面的链接建议使用`Array.push()` ,所以让我们首先创建一个新的数组来存储我们很快会有这样的小数组:
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```javascript
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var newArray = [];
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```
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> _现在尝试解决问题_
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## ![:speech_balloon:](https://forum.freecodecamp.com/images/emoji/emoji_one/speech_balloon.png?v=3 ":speech_balloon:")提示:2
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接下来我们需要一个`for loop`来遍历`arr` 。
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> _现在尝试解决问题_
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## ![:speech_balloon:](https://forum.freecodecamp.com/images/emoji/emoji_one/speech_balloon.png?v=3 ":speech_balloon:")提示:3
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最后,我们需要一个方法来进行实际的拆分,我们可以使用`Array.slice()`来做到这一点。这个算法的关键是理解`for loop` , `size` , `Array.slice()`和`Array.push()`是如何一起工作的。
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> _现在尝试解决问题_
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## 扰流警报!
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![警告牌](//discourse-user-assets.s3.amazonaws.com/original/2X/2/2d6c412a50797771301e7ceabd554cef4edcd74d.gif)
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**提前解决!**
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## ![:beginner:](https://forum.freecodecamp.com/images/emoji/emoji_one/beginner.png?v=3 ":初学者:")基本代码解决方案
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```javascript
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function chunkArrayInGroups(arr, size) {
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var temp = [];
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var result = [];
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for (var a = 0; a < arr.length; a++) {
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if (a % size !== size - 1)
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temp.push(arr[a]);
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else {
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temp.push(arr[a]);
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result.push(temp);
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temp = [];
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}
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}
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if (temp.length !== 0)
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result.push(temp);
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return result;
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}
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```
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![:rocket:](https://forum.freecodecamp.com/images/emoji/emoji_one/rocket.png?v=3 ":火箭:") [运行代码](https://repl.it/CLjU/24)
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### 代码说明:
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* 首先,我们创建了两个名为`temp`和`result`空数组,我们最终会返回它们。
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* 我们的**for循环**循环直到`a`等于或大于我们测试中数组的长度。
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* 在我们的循环中,我们使用`temp.push(arr[a]);`推送到`temp` `temp.push(arr[a]);`如果`a / size`的余数不等于`size - 1` 。
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* 否则,我们推送到`temp` ,将`temp`推送到`result`变量并将`temp`重置为空数组。
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* 接下来,如果`temp`不是空数组,我们将其推送到`result` 。
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* 最后,我们返回`result`的值。
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#### 相关链接
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* [的Array.push()](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/push)
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* [对于循环](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Statements/for)
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## ![:sunflower:](https://forum.freecodecamp.com/images/emoji/emoji_one/sunflower.png?v=3 ":向日葵:")中级代码解决方案:
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```javascript
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function chunkArrayInGroups(arr, size) {
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// Break it up.
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var arr2 = [];
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for (var i = 0; i < arr.length; i+=size) {
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arr2.push(arr.slice(i , i+size));
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}
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return arr2;
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}
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```
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![:rocket:](https://forum.freecodecamp.com/images/emoji/emoji_one/rocket.png?v=3 ":火箭:") [运行代码](https://repl.it/Cj9x/3)
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### 代码说明:
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* 首先,我们创建一个空数组`arr2` ,我们将存储'chunk'。
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* for循环从零开始,每次循环都按`size`递增,并在达到`arr.length`时停止。
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* 请注意,此for循环不会循环遍历`arr` 。相反,我们使用循环生成数字,我们可以将其用作索引,以在正确的位置切割数组。
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* 在我们的循环中,我们使用`arr.slice(i, i+size)`创建每个块,并使用`arr2.push()`将此值添加到`arr2` 。
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* 最后,我们返回`arr2`的值。
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#### 相关链接
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* [的Array.push()](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/push)
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* [Array.slice()](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/slice)
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* [对于循环](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Statements/for)
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## ![:rotating_light:](https://forum.freecodecamp.com/images/emoji/emoji_one/rotating_light.png?v=3 ":rotating_light:")高级代码解决方案
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```javascript
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function chunkArrayInGroups(arr, size) {
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// Break it up.
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var newArr = [];
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var i = 0;
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while (i < arr.length) {
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newArr.push(arr.slice(i, i+size));
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i += size;
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}
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return newArr;
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}
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chunkArrayInGroups(["a", "b", "c", "d"], 2);
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```
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![:rocket:](https://forum.freecodecamp.com/images/emoji/emoji_one/rocket.png?v=3 ":火箭:") [运行代码](https://repl.it/CLjU/26)
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### 代码说明:
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* 首先,我们创建两个变量。 `newArr`是一个空数组,我们将推送到。我们还将`i`变量设置为零,以便在我们的while循环中使用。
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* 我们的while循环循环,直到`i`等于或大于我们测试中数组的长度。
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* 在我们的循环中,我们使用`arr.slice(i, i+size)`推送到`newArr`数组。它第一次循环,它看起来像:
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newArr.push(arr.slice(1,1 + 2))
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* 在我们推送到`newArr` ,我们将`size`的变量添加到`i` 。
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* 最后,我们返回`newArr`的值。
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#### 相关链接
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* [的Array.push()](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/push)
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* [Array.slice()](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/slice)
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* [循环时](https://developer.mozilla.org/en/docs/Web/JavaScript/Reference/Statements/while)
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## ![:rotating_light:](https://forum.freecodecamp.com/images/emoji/emoji_one/rotating_light.png?v=3 ":rotating_light:")高级代码解决方案2:
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```javascript
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function chunkArrayInGroups(arr, size) {
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var newArr = [];
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while (arr.length) {
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newArr.push(arr.splice(0,size));
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}
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return newArr;
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}
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```
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![:rocket:](https://forum.freecodecamp.com/images/emoji/emoji_one/rocket.png?v=3 ":火箭:") [运行代码](https://repl.it/CLjU/579)
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### 代码说明:
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* 首先,我们创建一个变量。 `newArr`是一个空数组,我们将推送到。
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* 我们的`while`循环循环,直到我们测试中数组的长度不为0。
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* 在我们的循环中,我们使用`arr.splice(0, size)`推送到`newArr`数组。
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* 对于`while`循环的每次迭代,它会从`arr`前面删除元素的`size`数量,并将它们作为数组推送到`newArr` 。
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* 最后,我们返回`newArr`的值。
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#### 相关链接
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* [的Array.push()](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/push)
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* [方法Array.splice()](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/splice)
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* [循环时](https://developer.mozilla.org/en/docs/Web/JavaScript/Reference/Statements/while)
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## ![:rotating_light:](https://forum.freecodecamp.com/images/emoji/emoji_one/rotating_light.png?v=3 ":rotating_light:")高级代码解决方案3:
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```javascript
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function chunkArrayInGroups(arr, size) {
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if (arr.length <= size){
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return [arr];
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}
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else {
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return [arr.slice(0,size)].concat(chunkArrayInGroups(arr.slice(size),size));
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}
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}
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```
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![:rocket:](https://forum.freecodecamp.com/images/emoji/emoji_one/rocket.png?v=3 ":火箭:") [运行代码](https://repl.it/CLjU/579)
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### 代码说明:
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* 嵌套返回小于size的数组。
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* 对于任何大于大小的数组,它都被分成两部分。第一个段嵌套并与第二个第二段连接,进行递归调用。
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#### 相关链接
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* [递归](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Guide/Functions#Recursion)
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* [方法Array.splice()](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/splice)
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## ![:clipboard:](https://forum.freecodecamp.com/images/emoji/emoji_one/clipboard.png?v=3 ":剪贴板:")捐款说明:
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* ![:warning:](https://forum.freecodecamp.com/images/emoji/emoji_one/warning.png?v=3 ":警告:") **请勿**添加与任何现有解决方案类似的解决方案。如果您认为它**_相似但更好_** ,那么尝试合并(或替换)现有的类似解决方案。
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* 添加解决方案的说明。
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* 将解决方案分为以下类别之一 - **基本** , **中级**和**高级** 。 ![:traffic_light:](https://forum.freecodecamp.com/images/emoji/emoji_one/traffic_light.png?v=3 ":红绿灯:")
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* 如果您添加了任何**相关的主要内容,**请仅添加您的用户名。 ( ![:warning:](https://forum.freecodecamp.com/images/emoji/emoji_one/warning.png?v=3 ":警告:") **_不要_** _删除任何现有的用户名_ )
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> 看到![:point_right:](https://forum.freecodecamp.com/images/emoji/emoji_one/point_right.png?v=3 ":point_right:") [**`Wiki Challenge Solution Template`**](http://forum.freecodecamp.com/t/algorithm-article-template/14272)供参考。
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