b3213fc892
* fix: Chinese test suite Add localeTiltes, descriptions, and adjust test text and testStrings to get the automated test suite working. * fix: ran script, updated testStrings and solutions
88 lines
2.7 KiB
Markdown
88 lines
2.7 KiB
Markdown
---
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id: 587d8253367417b2b2512c6c
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title: Perform a Union on Two Sets
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challengeType: 1
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videoUrl: ''
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localeTitle: 在两个集上执行联合
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---
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## Description
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<section id="description">在本练习中，我们将对两组数据执行联合。我们将在我们的<code>Set</code>数据结构上创建一个名为<code>union</code> 。此方法应将另一个<code>Set</code>作为参数，并返回两个集合的<code>union</code>集，不包括任何重复值。例如，如果<code>setA = ['a','b','c']</code>和<code>setB = ['a','b','d','e']</code> ，则setA和setB的并集为： <code>setA.union(setB) = ['a', 'b', 'c', 'd', 'e']</code> 。 </section>
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## Instructions
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<section id="instructions">
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</section>
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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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- text: 你的<code>Set</code>类应该有一个<code>union</code>方法。
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testString: assert((function(){var test = new Set(); return (typeof test.union === 'function')})());
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- text: 收回了适当的收藏
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testString: 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 unionSetAB = setA.union(setB); var final = unionSetAB.values(); return (final.indexOf('a') !== -1 && final.indexOf('b') !== -1 && final.indexOf('c') !== -1 && final.indexOf('d') !== -1 && final.length === 4)})());
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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 Set() {
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// the var collection will hold the set
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var collection = [];
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// this method will check for the presence of an element and return true or false
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this.has = function(element) {
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return (collection.indexOf(element) !== -1);
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};
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// this method will return all the values in the set
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this.values = function() {
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return collection;
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};
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// this method will add an element to the set
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this.add = function(element) {
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if(!this.has(element)){
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collection.push(element);
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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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this.remove = function(element) {
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if(this.has(element)){
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var index = collection.indexOf(element);
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collection.splice(index,1);
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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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this.size = function() {
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return collection.length;
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};
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// change code below this line
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// change code above this line
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}
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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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```js
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// solution required
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```
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</section>
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