2018-10-10 22:03:03 +00:00
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---
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id: 56533eb9ac21ba0edf2244e1
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2021-03-15 03:20:39 +00:00
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title: 循环嵌套
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2018-10-10 22:03:03 +00:00
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challengeType: 1
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2020-04-29 10:29:13 +00:00
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videoUrl: 'https://scrimba.com/c/cRn6GHM'
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forumTopicId: 18248
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2021-01-13 02:31:00 +00:00
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dashedName: nesting-for-loops
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2018-10-10 22:03:03 +00:00
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---
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2020-12-16 07:37:30 +00:00
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# --description--
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2021-03-15 03:20:39 +00:00
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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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2020-04-29 10:29:13 +00:00
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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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2021-03-15 03:20:39 +00:00
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这里一次输出了 `arr` 中的每个子元素。 提示,对于内部循环,我们可以通过 `arr[i]` 的 `.length` 来获得子数组的长度,因为 `arr[i]` 本身就是一个数组。
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2018-10-10 22:03:03 +00:00
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2020-12-16 07:37:30 +00:00
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# --instructions--
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2018-10-10 22:03:03 +00:00
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2021-03-15 03:20:39 +00:00
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修改函数 `multiplyAll`,获得 `arr` 内部数组的每个数字相乘的结果 product。
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2018-10-10 22:03:03 +00:00
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2020-12-16 07:37:30 +00:00
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# --hints--
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2018-10-10 22:03:03 +00:00
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2021-03-15 03:20:39 +00:00
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`multiplyAll([[1],[2],[3]])` 应该返回 `6`
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2018-10-10 22:03:03 +00:00
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```js
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2020-12-16 07:37:30 +00:00
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assert(multiplyAll([[1], [2], [3]]) === 6);
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2018-10-10 22:03:03 +00:00
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```
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2021-03-15 03:20:39 +00:00
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`multiplyAll([[1,2],[3,4],[5,6,7]])` 应该返回 `5040`
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2018-10-10 22:03:03 +00:00
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2020-12-16 07:37:30 +00:00
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```js
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assert(
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multiplyAll([
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[1, 2],
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[3, 4],
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[5, 6, 7]
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]) === 5040
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);
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```
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2018-10-10 22:03:03 +00:00
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2021-03-15 03:20:39 +00:00
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`multiplyAll([[5,1],[0.2, 4, 0.5],[3, 9]])` 应该返回 `54`
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2020-04-29 10:29:13 +00:00
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2018-10-10 22:03:03 +00:00
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```js
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2020-12-16 07:37:30 +00:00
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assert(
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multiplyAll([
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[5, 1],
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[0.2, 4, 0.5],
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[3, 9]
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]) === 54
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);
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2018-10-10 22:03:03 +00:00
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```
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2020-04-29 10:29:13 +00:00
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2021-01-13 02:31:00 +00:00
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# --seed--
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## --seed-contents--
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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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multiplyAll([[1,2],[3,4],[5,6,7]]);
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```
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2020-12-16 07:37:30 +00:00
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# --solutions--
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2021-01-13 02:31:00 +00:00
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```js
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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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```
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