2018-10-10 22:03:03 +00:00
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---
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id: 587d8253367417b2b2512c6c
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2020-12-16 07:37:30 +00:00
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title: 在两个集上执行联合
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2018-10-10 22:03:03 +00:00
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challengeType: 1
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videoUrl: ''
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2021-01-13 02:31:00 +00:00
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dashedName: perform-a-union-on-two-sets
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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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2018-10-10 22:03:03 +00:00
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2020-12-16 07:37:30 +00:00
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在本练习中,我们将对两组数据执行联合。我们将在我们的`Set`数据结构上创建一个名为`union` 。此方法应将另一个`Set`作为参数,并返回两个集合的`union`集,不包括任何重复值。例如,如果`setA = ['a','b','c']`和`setB = ['a','b','d','e']` ,则setA和setB的并集为: `setA.union(setB) = ['a', 'b', 'c', 'd', 'e']` 。
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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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2020-12-16 07:37:30 +00:00
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你的`Set`类应该有一个`union`方法。
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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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(function () {
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var test = new Set();
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return typeof test.union === 'function';
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})()
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);
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2018-10-10 22:03:03 +00:00
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```
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2020-12-16 07:37:30 +00:00
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收回了适当的收藏
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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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(function () {
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var setA = new Set();
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var setB = new Set();
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setA.add('a');
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setA.add('b');
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setA.add('c');
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setB.add('c');
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setB.add('d');
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var unionSetAB = setA.union(setB);
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var final = unionSetAB.values();
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return (
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final.indexOf('a') !== -1 &&
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final.indexOf('b') !== -1 &&
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final.indexOf('c') !== -1 &&
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final.indexOf('d') !== -1 &&
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final.length === 4
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);
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})()
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);
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2018-10-10 22:03:03 +00:00
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```
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2020-08-13 15:24:35 +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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class Set {
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constructor() {
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// This will hold the set
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this.dictionary = {};
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this.length = 0;
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}
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// This method will check for the presence of an element and return true or false
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has(element) {
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return this.dictionary[element] !== undefined;
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}
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// This method will return all the values in the set
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values() {
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return Object.keys(this.dictionary);
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}
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// This method will add an element to the set
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add(element) {
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if (!this.has(element)) {
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this.dictionary[element] = true;
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this.length++;
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return true;
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}
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return false;
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}
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// This method will remove an element from a set
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remove(element) {
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if (this.has(element)) {
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delete this.dictionary[element];
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this.length--;
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return true;
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}
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return false;
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}
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// This method will return the size of the set
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size() {
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return this.length;
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}
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// Only change code below this line
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// Only change code above this line
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}
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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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class Set {
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constructor() {
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this.dictionary = {};
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this.length = 0;
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}
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has(element) {
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return this.dictionary[element] !== undefined;
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}
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values() {
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return Object.keys(this.dictionary);
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}
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add(element) {
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if (!this.has(element)) {
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this.dictionary[element] = true;
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this.length++;
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return true;
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}
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return false;
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}
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remove(element) {
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if (this.has(element)) {
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delete this.dictionary[element];
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this.length--;
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return true;
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}
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return false;
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}
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size() {
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return this.length;
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}
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union(set) {
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const newSet = new Set();
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this.values().forEach(value => {
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newSet.add(value);
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})
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set.values().forEach(value => {
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newSet.add(value);
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})
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return newSet;
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}
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}
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```
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