JavaScript 求两个数组的交集，并集，差集，去重

a = [1, 2, 3]，b = [2, 4, 5]

1.差集

（a-b 差集：属于a但不属于b的集合）a-b=[1,3],b-a=[4,5]

• filter + includes

// 差集
let difference = a.concat(b).filter(v => !a.includes(v) ) 

• Set + Array.from

let aSet = new Set(a)
let bSet = new Set(b)
let difference = Array.from(new Set(a.concat(b).filter(v => !aSet.has(v) ))) 

• filter + indexOf

var difference = a.filter(function(v){ return b.indexOf(v) === -1 })

2.交集

• filter + includes

// 交集
let intersection = a.filter(v => b.includes(v)) // [2]

• Set + Array.from

let aSet = new Set(a)
let bSet = new Set(b)
let intersection = Array.from(new Set(a.filter(v => bSet.has(v)))) // [2]

• filter + indexOf

// 交集
var intersection = a.filter(function(v){ return b.indexOf(v) > -1 }) // [2]

3.并集

• filter + includes

let union = a.concat(b.filter(v => !a.includes(v))) // [1,2,3,4,5]

• Set + Array.from

let aSet = new Set(a)
let bSet = new Set(b)
let union = Array.from(new Set(a.concat(b))) // [1,2,3,4,5]

• filter + indexOf

var union = a.concat(b.filter(function(v) {return a.indexOf(v) === -1 } )) // [1,2,3,4,5]

4.去重

• 双层循环，外层循环元素，内层循环时比较值，如果有相同的值则跳过，不相同则push进数组

Array.prototype.distinct = function(){
 var arr = this,
  result = [],
  i,
  j,
  len = arr.length;
 for(i = 0; i < len; i++){
  for(j = i + 1; j < len; j++){
   if(arr[i] === arr[j]){
    j = ++i;
   }
  }
  result.push(arr[i]);
 }
 return result;
}
var arra = [1,2,3,4,4,1,1,2,1,1,1];
arra.distinct();    //返回[3,4,2,1]

• 利用splice直接在原数组进行操作
  双层循环，外层循环元素，内层循环时比较值
  值相同时，则删去这个值
  注意点:删除元素之后，需要将数组的长度也减1.

Array.prototype.distinct = function (){
 var arr = this,
  i,
  j,
  len = arr.length;
 for(i = 0; i < len; i++){
  for(j = i + 1; j < len; j++){
   if(arr[i] == arr[j]){
    arr.splice(j,1);
    len--;
    j--;
   }
  }
 }
 return arr;
};
var a = [1,2,3,4,5,6,5,3,2,4,56,4,1,2,1,1,1,1,1,1,];
var b = a.distinct();
console.log(b.toString()); //1,2,3,4,5,6,56

• 利用对象的属性不能相同的特点进行去重

Array.prototype.distinct = function (){
 var arr = this,
  i,
  obj = {},
  result = [],
  len = arr.length;
 for(i = 0; i< arr.length; i++){
  if(!obj[arr[i]]){ //如果能查找到，证明数组元素重复了
   obj[arr[i]] = 1;
   result.push(arr[i]);
  }
 }
 return result;
};
var a = [1,2,3,4,5,6,5,3,2,4,56,4,1,2,1,1,1,1,1,1,];
var b = a.distinct();
console.log(b.toString()); //1,2,3,4,5,6,56

• 数组递归去重
  运用递归的思想
  先排序，然后从最后开始比较，遇到相同，则删除

Array.prototype.distinct = function (){
 var arr = this,
  len = arr.length;
 arr.sort(function(a,b){  //对数组进行排序才能方便比较
  return a - b;
 })
 function loop(index){
  if(index >= 1){
   if(arr[index] === arr[index-1]){
    arr.splice(index,1);
   }
   loop(index - 1); //递归loop函数进行去重
  }
 }
 loop(len-1);
 return arr;
};
var a = [1,2,3,4,5,6,5,3,2,4,56,4,1,2,1,1,1,1,1,1,56,45,56];
var b = a.distinct();
console.log(b.toString());  //1,2,3,4,5,6,45,56

• 利用indexOf以及forEach

Array.prototype.distinct = function (){
 var arr = this,
  result = [],
  len = arr.length;
 arr.forEach(function(v, i ,arr){  //这里利用map，filter方法也可以实现
  var bool = arr.indexOf(v,i+1);  //从传入参数的下一个索引值开始寻找是否存在重复
  if(bool === -1){
   result.push(v);
  }
 })
 return result;
};
var a = [1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,3,3,3,3,3,3,3,2,3,3,2,2,1,23,1,23,2,3,2,3,2,3];
var b = a.distinct();
console.log(b.toString()); //1,23,2,3

• 利用ES6的set
  Set数据结构，它类似于数组，其成员的值都是唯一的。

利用Array.from将Set结构转换成数组
function dedupe(array){
 return Array.from(new Set(array));
}
dedupe([1,1,2,3]) //[1,2,3]

拓展运算符(...)内部使用for...of循环
let arr = [1,2,3,3];
let resultarr = [...new Set(arr)]; 
console.log(resultarr); //[1,2,3]

参考：https://excaliburhan.com/post/js-set-operation.html
https://www.jb51.net/article/118657.htm