分别写出数组的交集、并集、差集、补集这四个方法

const intersect = (a, b) => a.filter(i => b.includes(i)) // 交const exclude = (a, b) => a.filter(i => !b.includes(i)) // 差const union = (a, b) => exclude(a, b).concat(b) // 并const unionAll = (a, b) => a.concat(b) // 重复并const xor = (a, b) => exclude(a, b).concat(exclude(b, a)) // 补

这几个方法全是 O(n2) 的复杂度…性能很差