2018-10-10 18:03:03 -04:00
|
|
|
|
---
|
|
|
|
|
id: 587d8253367417b2b2512c6d
|
2020-12-16 00:37:30 -07:00
|
|
|
|
title: 在两组数据上执行交集
|
2018-10-10 18:03:03 -04:00
|
|
|
|
challengeType: 1
|
|
|
|
|
videoUrl: ''
|
|
|
|
|
---
|
|
|
|
|
|
2020-12-16 00:37:30 -07:00
|
|
|
|
# --description--
|
2018-10-10 18:03:03 -04:00
|
|
|
|
|
2020-12-16 00:37:30 -07:00
|
|
|
|
在本练习中,我们将对两组数据执行交集。我们将在我们的`Set`数据结构上创建一个名为`intersection` 。集合的交集表示两个或更多集合共有的所有值。此方法应将另一个`Set`作为参数,并返回两个集合的`intersection` 。例如,如果`setA = ['a','b','c']`和`setB = ['a','b','d','e']` ,则setA和setB的交集为: `setA.intersection(setB) = ['a', 'b']` 。
|
2018-10-10 18:03:03 -04:00
|
|
|
|
|
2020-12-16 00:37:30 -07:00
|
|
|
|
# --hints--
|
2018-10-10 18:03:03 -04:00
|
|
|
|
|
2020-12-16 00:37:30 -07:00
|
|
|
|
您的`Set`类应该有一个`intersection`方法。
|
2018-10-10 18:03:03 -04:00
|
|
|
|
|
|
|
|
|
```js
|
2020-12-16 00:37:30 -07:00
|
|
|
|
assert(
|
|
|
|
|
(function () {
|
|
|
|
|
var test = new Set();
|
|
|
|
|
return typeof test.intersection === 'function';
|
|
|
|
|
})()
|
|
|
|
|
);
|
2018-10-10 18:03:03 -04:00
|
|
|
|
```
|
|
|
|
|
|
2020-12-16 00:37:30 -07:00
|
|
|
|
收回了适当的收藏
|
2018-10-10 18:03:03 -04:00
|
|
|
|
|
|
|
|
|
```js
|
2020-12-16 00:37:30 -07:00
|
|
|
|
assert(
|
|
|
|
|
(function () {
|
|
|
|
|
var setA = new Set();
|
|
|
|
|
var setB = new Set();
|
|
|
|
|
setA.add('a');
|
|
|
|
|
setA.add('b');
|
|
|
|
|
setA.add('c');
|
|
|
|
|
setB.add('c');
|
|
|
|
|
setB.add('d');
|
|
|
|
|
var intersectionSetAB = setA.intersection(setB);
|
|
|
|
|
return (
|
|
|
|
|
intersectionSetAB.size() === 1 && intersectionSetAB.values()[0] === 'c'
|
|
|
|
|
);
|
|
|
|
|
})()
|
|
|
|
|
);
|
2018-10-10 18:03:03 -04:00
|
|
|
|
```
|
2020-08-13 17:24:35 +02:00
|
|
|
|
|
2020-12-16 00:37:30 -07:00
|
|
|
|
# --solutions--
|
|
|
|
|
|