不适用set做法
var a = [1,2,3,4,5]
var b = [2,4,6,8,10]
var c = a.filter(function(v){ return b.indexOf(v) > -1 })
var d = a.filter(function(v){ return b.indexOf(v) == -1 })
var e = a.filter(function(v){ return !(b.indexOf(v) > -1) })
.concat(b.filter(function(v){ return !(a.indexOf(v) > -1)}))
var f = a.concat(b.filter(function(v){ return !(a.indexOf(v) > -1)}));
console.log("数组a:", a);
console.log("数组b:", b);
console.log("a与b的交集:", c);
console.log("a与b的差集:", d);
console.log("a与b的补集:", e);
console.log("a与b的并集:", f);
使用set的做法
var a = [1,2,3,4,5]
var b = [2,4,6,8,10]
console.log("数组a:", a);
console.log("数组b:", b);
var sa = new Set(a);
var sb = new Set(b);
let intersect = a.filter(x => sb.has(x));
let minus = a.filter(x => !sb.has(x));
let complement = [...a.filter(x => !sb.has(x)), ...b.filter(x => !sa.has(x))];
let unionSet = Array.from(new Set([...a, ...b]));
[...new Set([...a, ...b])]
console.log("a与b的交集:", intersect);
console.log("a与b的差集:", minus);
console.log("a与b的补集:", complement);
console.log("a与b的并集:", unionSet);