5.3 KiB
5.3 KiB
id | title | challengeType | videoUrl | localeTitle |
---|---|---|---|---|
587d8254367417b2b2512c6f | Perform a Subset Check on Two Sets of Data | 1 | 对两组数据执行子集检查 |
Description
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']
和setB = ['a','b','c','d']
,则setA和setB的子集为: setA.subset(setB)
应该是true
。 Instructions
Tests
tests:
- text: 你的<code>Set</code>类应该有一个<code>union</code>方法。
testString: 'assert(function(){var test = new Set(); return (typeof test.subset === "function")}, "Your <code>Set</code> class should have a <code>union</code> method.");'
- 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)}, "The first Set() was contained in the second Set");'
- text: '<code>["a", "b"].subset(["a", "b", "c", "d"])</code>应该返回<code>true</code> “)'
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)}, "<code>["a", "b"].subset(["a", "b", "c", "d"])</code> should return <code>true</code>");'
- text: '<code>["a", "b", "c"].subset(["a", "b"])</code>应返回<code>false</code> “)'
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)}, "<code>["a", "b", "c"].subset(["a", "b"])</code> should return <code>false</code>");'
- text: '<code>[].subset([])</code>应该返回<code>true</code>'
testString: 'assert(function(){var setA = new Set(); var setB = new Set(); var subsetSetAB = setA.subset(setB); return (subsetSetAB === true)}, "<code>[].subset([])</code> should return <code>true</code>");'
- text: '<code>["a", "b"].subset(["c", "d"])</code>应返回<code>false</code> “)'
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)}, "<code>["a", "b"].subset(["c", "d"])</code> should return <code>false</code>");'
Challenge Seed
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
// solution required