freeCodeCamp/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/perform-a-subset-check-on-t...

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---
id: 587d8254367417b2b2512c6f
title: 对两组数据执行子集检查
challengeType: 1
videoUrl: ''
dashedName: perform-a-subset-check-on-two-sets-of-data
---
# --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']`和`setB = ['a','b','c','d']` 则setA和setB的子集为 `setA.subset(setB)`应该是`true` 。
# --hints--
你的`Set`类应该有一个`union`方法。
```js
assert(
(function () {
var test = new Set();
return typeof test.subset === 'function';
})()
);
```
第一个Set包含在第二个Set中
```js
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;
})()
);
```
`["a", "b"].subset(["a", "b", "c", "d"])`应该返回`true` “)
```js
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;
})()
);
```
`["a", "b", "c"].subset(["a", "b"])`应返回`false` “)
```js
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;
})()
);
```
`[].subset([])`应该返回`true`
```js
assert(
(function () {
var setA = new Set();
var setB = new Set();
var subsetSetAB = setA.subset(setB);
return subsetSetAB === true;
})()
);
```
`["a", "b"].subset(["c", "d"])`应返回`false` “)
```js
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;
})()
);
```
# --seed--
## --seed-contents--
```js
class Set {
constructor() {
// This will hold the set
this.dictionary = {};
this.length = 0;
}
// This method will check for the presence of an element and return true or false
has(element) {
return this.dictionary[element] !== undefined;
}
// This method will return all the values in the set
values() {
return Object.keys(this.dictionary);
}
// This method will add an element to the set
add(element) {
if (!this.has(element)) {
this.dictionary[element] = true;
this.length++;
return true;
}
return false;
}
// This method will remove an element from a set
remove(element) {
if (this.has(element)) {
delete this.dictionary[element];
this.length--;
return true;
}
return false;
}
// This method will return the size of the set
size() {
return this.length;
}
// This is our union method
union(set) {
const newSet = new Set();
this.values().forEach(value => {
newSet.add(value);
})
set.values().forEach(value => {
newSet.add(value);
})
return newSet;
}
// This is our intersection method
intersection(set) {
const newSet = new Set();
let largeSet;
let smallSet;
if (this.dictionary.length > set.length) {
largeSet = this;
smallSet = set;
} else {
largeSet = set;
smallSet = this;
}
smallSet.values().forEach(value => {
if (largeSet.dictionary[value]) {
newSet.add(value);
}
})
return newSet;
}
difference(set) {
const newSet = new Set();
this.values().forEach(value => {
if (!set.dictionary[value]) {
newSet.add(value);
}
})
return newSet;
}
// Only change code below this line
// Only change code above this line
}
```
# --solutions--
```js
class Set {
constructor() {
this.dictionary = {};
this.length = 0;
}
has(element) {
return this.dictionary[element] !== undefined;
}
values() {
return Object.keys(this.dictionary);
}
add(element) {
if (!this.has(element)) {
this.dictionary[element] = true;
this.length++;
return true;
}
return false;
}
remove(element) {
if (this.has(element)) {
delete this.dictionary[element];
this.length--;
return true;
}
return false;
}
size() {
return this.length;
}
union(set) {
const newSet = new Set();
this.values().forEach(value => {
newSet.add(value);
})
set.values().forEach(value => {
newSet.add(value);
})
return newSet;
}
intersection(set) {
const newSet = new Set();
let largeSet;
let smallSet;
if (this.dictionary.length > set.length) {
largeSet = this;
smallSet = set;
} else {
largeSet = set;
smallSet = this;
}
smallSet.values().forEach(value => {
if (largeSet.dictionary[value]) {
newSet.add(value);
}
})
return newSet;
}
difference(set) {
const newSet = new Set();
this.values().forEach(value => {
if (!set.dictionary[value]) {
newSet.add(value);
}
})
return newSet;
}
isSubsetOf(set) {
for(const value of this.values()){
if(!set.dictionary[value]) return false;
}
return true
}
}
```