一、简单数组
1、ES5：

const arr1 = [1,2,3,4,5],
      arr2 = [5,6,7,8,9];

// 交集
let intersection = arr1.filter(function (val) { return arr2.indexOf(val) > -1 })

// 并集
let union = arr1.concat(arr2.filter(function (val) { return !(arr1.indexOf(val) > -1) }))

// 补集 两个数组各自没有的集合
let complement = arr1.filter(function (val) { return !(arr2.indexOf(val) > -1) })
.concat(arr2.filter(function (val) { return !(arr1.indexOf(val) > -1) }))

// 差集 数组arr1相对于arr2所没有的
let diff = arr1.filter(function (val) { return arr2.indexOf(val) === -1 })

console.log('arr1: ', arr1);
console.log('arr2: ', arr2);
console.log('交集', intersection);
console.log('并集', union);
console.log('补集', complement);
console.log('差集', diff);

ES6：

const arr1 = [1,2,3,4,5],
      arr2 = [5,6,7,8,9],
      _arr1Set = new Set(arr1),
      _arr2Set = new Set(arr2);


// 交集
let intersection = arr1.filter(item => _arr2Set.has(item))

// 并集
let union = Array.from(new Set([...arr1, ...arr2]))

// 补集 两个数组各自没有的集合
let complement = [...arr1.filter(item => !_arr2Set.has(item)), ...arr2.filter(item => !_arr1Set.has(item))]

// 差集 数组arr1相对于arr2所没有的
let diff = arr1.filter(item => !_arr2Set.has(item))
console.log('arr1: ', arr1);
console.log('arr2: ', arr2);
console.log('交集', intersection);
console.log('并集', union);
console.log('补集', complement);
console.log('差集', diff);

JQ:

const arr1 = [1,2,3,4,5],
      arr2 = [5,6,7,8,9];


// 交集
let intersection = $(arr1).filter(arr2).toArray();

// 并集
let union = $.unique(arr1.concat(arr2))

// 补集 两个数组各自没有的集合
let complement = $(arr1).not(arr2).toArray().concat($(arr2).not(arr1).toArray())

// 差集 数组arr1相对于arr2所没有的
let diff = $(arr1).not(arr2).toArray()
console.log('arr1: ', arr1);
console.log('arr2: ', arr2);
console.log('交集', intersection);
console.log('并集', union);
console.log('补集', complement);
console.log('差集', diff);

二、对象数组

// 形如如下数组
let arr1 = [], arr2 = [];
arr1 = [
  {
    ID: 1,
    Name: 1,
    desc: 'Number'
  },
  {
    ID: 2,
    Name: 2,
    desc: 'Number'
  },
  {
    ID: 3,
    Name: 3,
    desc: 'Number'
  },
  {
    ID: 4,
    Name: 4,
    desc: 'Number'
  },
  {
    ID: 5,
    Name: 5,
    desc: 'Number'
  }
]
arr2 = [
{
    ID: 5,
    Name: 5,
    desc: 'Number'
  },
  {
    ID: 6,
    Name: 6,
    desc: 'Number'
  },
  {
    ID: 7,
    Name: 7,
    desc: 'Number'
  },
  {
    ID: 8,
    Name: 8,
    desc: 'Number'
  },
  {
    ID: 9,
    Name: 9,
    desc: 'Number'
  }
]

// 交集
let intersection = []
for (let i = 0, len = arr1.length; i < len; i++) {
  for (let j = 0, length = arr2.length; j < length; j++) {
    if (arr1[i].ID === arr2[j].ID) {
      intersection.push(arr1[i])
    }
  }
}
console.log('交集', intersection)

// 并集
let union = [...arr1, ...arr2]
for (let i = 0, len = arr1.length; i < len; i++ ) {
  for (let j = 0, length = arr2.length; j < length; j++) {
    if (arr1[i].ID === arr2[j].ID) {
      union.splice(union.findIndex(item => item.ID === arr1[i].ID), 1)
    }
  }
}
console.log('并集', union)

// 补集
let complement = [...arr1, ...arr2]
for (let i = 0, len = arr1.length; i < len; i++ ) {
  for (let j = 0, length = arr2.length; j < length; j++) {
    if (arr1[i].ID === arr2[j].ID) {
      complement.splice(complement.findIndex(item => item.ID === arr1[i].ID), 1)
      complement.splice(complement.findIndex(item => item.ID === arr2[j].ID), 1)
    }
  }
}
console.log('补集', complement)

// 差集
let diff = [...arr1]
for (let i = 0, len = arr1.length; i < len; i++ ) {
  let flag = false
  for (let j = 0, length = arr2.length; j < length; j++) {
    if (arr1[i].ID === arr2[j].ID) {
      flag = true
    }
  }
  if (flag) {
    diff.splice(diff.findIndex(item => item.ID === arr1[i].ID), 1)
  }
}
console.log('差集', diff)

其他更优雅的写法，待续……

更新 2021.9.27

针对道友飞扬_柳絮提出的一个问题：[1,1]，[1,1]取交集不正确的问题，这里做一下说明。
一般我们取两个数组的交集，不管是简单数组还是对象数组，都是先对数组进行一个去重，再进行取交集的操作，减少不必要的遍历。