2018-10-10 18:03:03 -04:00
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---
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id: 56533eb9ac21ba0edf2244e1
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challengeType: 1
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2020-04-29 18:29:13 +08:00
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videoUrl: 'https://scrimba.com/c/cRn6GHM'
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forumTopicId: 18248
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2020-10-01 17:54:21 +02:00
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title: 循环嵌套
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2018-10-10 18:03:03 -04:00
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---
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## Description
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2020-04-29 18:29:13 +08:00
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<section id='description'>
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如果你有一个二维数组,可以使用相同的逻辑,先遍历外面的数组,再遍历里面的子数组。下面是一个例子:
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```js
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var arr = [
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[1,2], [3,4], [5,6]
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];
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for (var i=0; i < arr.length; i++) {
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for (var j=0; j < arr[i].length; j++) {
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console.log(arr[i][j]);
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}
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}
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```
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一次输出<code>arr</code>中的每个子元素。提示,对于内部循环,我们可以通过<code>arr[i]</code>的<code>.length</code>来获得子数组的长度,因为<code>arr[i]</code>的本身就是一个数组。
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</section>
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2018-10-10 18:03:03 -04:00
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## Instructions
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2020-04-29 18:29:13 +08:00
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<section id='instructions'>
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修改函数<code>multiplyAll</code>,获得<code>arr</code>内部数组的每个数字相乘的结果<code>product</code>。
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</section>
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2018-10-10 18:03:03 -04:00
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## Tests
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<section id='tests'>
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```yml
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tests:
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2020-04-29 18:29:13 +08:00
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- text: <code>multiplyAll([[1],[2],[3]])</code>应该返回 <code>6</code>。
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2020-02-18 01:40:55 +09:00
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testString: assert(multiplyAll([[1],[2],[3]]) === 6);
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2020-04-29 18:29:13 +08:00
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- text: <code>multiplyAll([[1,2],[3,4],[5,6,7]])</code>应该返回 <code>5040</code>。
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2020-02-18 01:40:55 +09:00
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testString: assert(multiplyAll([[1,2],[3,4],[5,6,7]]) === 5040);
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2020-04-29 18:29:13 +08:00
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- text: <code>multiplyAll([[5,1],[0.2, 4, 0.5],[3, 9]])</code>应该返回 <code>54</code>。
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2020-02-18 01:40:55 +09:00
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testString: assert(multiplyAll([[5,1],[0.2, 4, 0.5],[3, 9]]) === 54);
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2018-10-10 18:03:03 -04:00
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```
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</section>
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## Challenge Seed
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<section id='challengeSeed'>
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<div id='js-seed'>
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```js
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function multiplyAll(arr) {
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var product = 1;
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// Only change code below this line
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// Only change code above this line
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return product;
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}
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// Modify values below to test your code
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multiplyAll([[1,2],[3,4],[5,6,7]]);
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```
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</div>
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</section>
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## Solution
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<section id='solution'>
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2020-04-29 18:29:13 +08:00
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2018-10-10 18:03:03 -04:00
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```js
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2020-04-29 18:29:13 +08:00
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function multiplyAll(arr) {
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var product = 1;
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for (var i = 0; i < arr.length; i++) {
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for (var j = 0; j < arr[i].length; j++) {
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product *= arr[i][j];
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}
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}
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return product;
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}
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multiplyAll([[1,2],[3,4],[5,6,7]]);
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2018-10-10 18:03:03 -04:00
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```
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2020-04-29 18:29:13 +08:00
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2018-10-10 18:03:03 -04:00
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</section>
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