2018-10-10 18:03:03 -04:00
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---
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id: a3566b1109230028080c9345
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2020-12-16 00:37:30 -07:00
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title: 范围内的数字求和
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2018-10-10 18:03:03 -04:00
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challengeType: 5
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2020-09-07 16:10:29 +08:00
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forumTopicId: 16083
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2021-01-13 03:31:00 +01:00
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dashedName: sum-all-numbers-in-a-range
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2018-10-10 18:03:03 -04:00
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---
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2020-12-16 00:37:30 -07:00
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# --description--
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2020-09-07 16:10:29 +08:00
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2021-01-12 08:18:51 -08:00
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给出一个含有两个数字的数组,我们需要写一个函数,让它返回这两个数字间所有数字(包含这两个数字)的总和。注意,较小数不一定总是出现在数组的第一个元素。
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2018-10-10 18:03:03 -04:00
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2021-01-12 08:18:51 -08:00
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例如,`sumAll([4,1])` 应返回 `10`,因为从 1 到 4(包含 1、4)的所有数字的和是 `10`。
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2018-10-10 18:03:03 -04:00
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2020-12-16 00:37:30 -07:00
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# --hints--
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2018-10-10 18:03:03 -04:00
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2021-01-12 08:18:51 -08:00
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`sumAll([1, 4])` 应返回一个数字。
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2018-10-10 18:03:03 -04:00
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2020-12-16 00:37:30 -07:00
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```js
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assert(typeof sumAll([1, 4]) === 'number');
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2018-10-10 18:03:03 -04:00
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```
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2021-01-12 08:18:51 -08:00
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`sumAll([1, 4])` 应返回 10。
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2018-10-10 18:03:03 -04:00
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```js
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2020-12-16 00:37:30 -07:00
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assert.deepEqual(sumAll([1, 4]), 10);
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2018-10-10 18:03:03 -04:00
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```
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2021-01-12 08:18:51 -08:00
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`sumAll([4, 1])` 应返回 10。
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2018-10-10 18:03:03 -04:00
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2020-12-16 00:37:30 -07:00
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```js
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assert.deepEqual(sumAll([4, 1]), 10);
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```
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2018-10-10 18:03:03 -04:00
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2021-01-12 08:18:51 -08:00
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`sumAll([5, 10])` 应返回 45。
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2018-10-10 18:03:03 -04:00
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2020-12-16 00:37:30 -07:00
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```js
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assert.deepEqual(sumAll([5, 10]), 45);
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```
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2018-10-10 18:03:03 -04:00
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2021-01-12 08:18:51 -08:00
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`sumAll([10, 5])` 应返回 45。
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2020-09-07 16:10:29 +08:00
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2018-10-10 18:03:03 -04:00
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```js
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2020-12-16 00:37:30 -07:00
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assert.deepEqual(sumAll([10, 5]), 45);
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2018-10-10 18:03:03 -04:00
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```
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2020-12-16 00:37:30 -07:00
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2021-01-13 03:31:00 +01:00
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# --seed--
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## --seed-contents--
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```js
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function sumAll(arr) {
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return 1;
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}
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sumAll([1, 4]);
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```
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2020-12-16 00:37:30 -07:00
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# --solutions--
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2021-01-13 03:31:00 +01:00
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```js
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function sumAll(arr) {
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var sum = 0;
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arr.sort(function(a,b) {return a-b;});
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for (var i = arr[0]; i <= arr[1]; i++) {
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sum += i;
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}
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return sum;
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}
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```
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