--- id: 587d8254367417b2b2512c6f title: Perform a Subset Check on Two Sets of Data challengeType: 1 videoUrl: '' localeTitle: 对两组数据执行子集检查 --- ## Description
在本练习中,我们将对2组数据执行子集测试。我们将在我们的Set数据结构上创建一个名为subset 。这将比较第一组与第二组,如果第一组完全包含在第二组中,则它将返回true。例如,如果setA = ['a','b']setB = ['a','b','c','d'] ,则setA和setB的子集为: setA.subset(setB)应该是true
## Instructions
## Tests
```yml tests: - text: 你的Set类应该有一个union方法。 testString: assert((function(){var test = new Set(); return (typeof test.subset === 'function')})()); - text: 第一个Set()包含在第二个Set中 testString: assert((function(){var setA = new Set(); var setB = new Set(); setA.add('a'); setB.add('b'); setB.add('c'); setB.add('a'); setB.add('d'); var subsetSetAB = setA.subset(setB);return (subsetSetAB === true)})()); - text: '["a", "b"].subset(["a", "b", "c", "d"])应该返回true “)' testString: assert((function(){var setA = new Set(); var setB = new Set(); setA.add('a'); setA.add('b'); setB.add('a'); setB.add('b'); setB.add('c'); setB.add('d'); var subsetSetAB = setA.subset(setB); return (subsetSetAB === true)})()); - text: '["a", "b", "c"].subset(["a", "b"])应返回false “)' testString: assert((function(){var setA = new Set(); var setB = new Set(); setA.add('a'); setA.add('b'); setA.add('c'); setB.add('a'); setB.add('b'); var subsetSetAB = setA.subset(setB); return (subsetSetAB === false)})()); - text: '[].subset([])应该返回true' testString: assert((function(){var setA = new Set(); var setB = new Set(); var subsetSetAB = setA.subset(setB); return (subsetSetAB === true)})()); - text: '["a", "b"].subset(["c", "d"])应返回false “)' testString: assert((function(){var setA = new Set(); var setB = new Set(); setA.add('a'); setA.add('b'); setB.add('c'); setB.add('d'); var subsetSetAB = setA.subset(setB); return (subsetSetAB === false)})()); ```
## Challenge Seed
```js function Set() { // the var collection will hold the set var collection = []; // this method will check for the presence of an element and return true or false this.has = function(element) { return (collection.indexOf(element) !== -1); }; // this method will return all the values in the set this.values = function() { return collection; }; // this method will add an element to the set this.add = function(element) { if(!this.has(element)){ collection.push(element); return true; } return false; }; // this method will remove an element from a set this.remove = function(element) { if(this.has(element)){ var index = collection.indexOf(element); collection.splice(index,1); return true; } return false; }; // this method will return the size of the collection this.size = function() { return collection.length; }; // this method will return the union of two sets this.union = function(otherSet) { var unionSet = new Set(); var firstSet = this.values(); var secondSet = otherSet.values(); firstSet.forEach(function(e){ unionSet.add(e); }); secondSet.forEach(function(e){ unionSet.add(e); }); return unionSet; }; // this method will return the intersection of two sets as a new set this.intersection = function(otherSet) { var intersectionSet = new Set(); var firstSet = this.values(); firstSet.forEach(function(e){ if(otherSet.has(e)){ intersectionSet.add(e); } }); return intersectionSet; }; // this method will return the difference of two sets as a new set this.difference = function(otherSet) { var differenceSet = new Set(); var firstSet = this.values(); firstSet.forEach(function(e){ if(!otherSet.has(e)){ differenceSet.add(e); } }); return differenceSet; }; // change code below this line // change code above this line } ```
## Solution
```js // solution required ```