数组交集，并集，差集

[Sunny-117]

No description provided.

[qiuye-zhou]

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

function Union () {
    const args = [].slice.call(arguments)
    const arrs = args.reduce((total,current) => {
        return total.concat(current)
    },[])
    return [...new Set(arrs)]
}
console.log('------')
// 返回传入所有数组的并集
console.log(Union(arr1, arr2))

function intersections () {
    let res
    const args = [].slice.call(arguments).map(e => new Set(e))
    args.slice(1).forEach(e => {
        res = [...new Set([...args[0]].filter(x => e.has(x)))]
    })
    return res
}
console.log('------')
// 返回传入所有数组的交集
console.log(intersections(arr1, arr2, [3]))

function subtraction () {
    const result = []
    let res = []
    const args = [].slice.call(arguments).map(e => new Set(e))
    args.forEach((e, ei) => {
        res = []
        args.forEach((x, xi) => {
            if (ei !== xi) {
                res.push([...new Set([...e].filter(y => !x.has(y)))])
            }
        })
        result.push(res)
    })
    return result
}
console.log('------')
// 返回传入的所有数组相对于其他数组的差集
console.log(subtraction(arr1, arr2, [9, 10, 11]))

[bearki99]

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

// 1. 并集
const arr3 = [...new Set(arr1.concat(arr2))];

// 2. 交集
const set = new Set(arr1);
const res = [];
for (let i = 0; i < arr2.length; i++) {
  if (set.has(arr2[i]) && res.indexOf(arr2[i]) === -1) {
    res.push(arr2[i]);
  }
}
console.log(res);

// 3. 差集 这里求的是除了交集之外的
const res2 = [...new Set(arr1.concat(arr2))];
for(let i = 0; i < res2.length; i++){
    if(res.indexOf(res2[i])!== -1) res2.splice(i, 1);
}
console.log(res2);

[veneno-o]

// 1. 并集
const res1 = [...new Set(arr1.concat(arr2))];
console.log(res1)

// 2. 交集
const set = new Set(arr1);
const res2 = [];
for (let i = 0; i < arr2.length; i++) {
  if (set.has(arr2[i]) && !res2.includes(arr2[i])) {
    res2.push(arr2[i]);
  }
}
console.log(res2);

// 3. 差集 这里求的是除了交集之外的
const res3 = res1.filter(item => !res2.includes(item))
console.log(res3);

[AaronKevinAA]

let arr1 = [1,2,3]
let arr2 = [2,3,4]

// 并集 解构+set+解构
let unionArr = [...new Set([...arr1,...arr2])]
console.log(unionArr) //[1,2,3,4]

// 交集
let res = arr1.filter(item=>{
return arr2.includes(item)
})
console.log(res)

// 差集 arr1-arr2
let ans = arr1.filter(item=>{
return !arr2.includes(item)
})
console.log(ans)

[kangkang123269]

let arr1 = [1, 2, 3, 4];
let arr2 = [3, 4, 5, 6];

// 并集
let union = [...new Set([...arr1,...arr2])];
console.log(union); // 输出：[1 ,2 ,3 ,4 ,5 ，6]

//交集
let intersection = arr1.filter(x => new Set(arr2).has(x));
console.log(intersection); // 输出：[3 ，4]

// 差集 (arr1 相对于 arr2 的差集)
let difference = arr1.filter(x => !new Set(arr2).has(x));
console.log(difference); // 输出：[1 ，2]

[huxuedong]

const setA = new Set([1, 2, 3]);
const setB = new Set([2, 3, 4]);

// 并集
const union = new Set([...setA, ...setB]);
console.log(union); // Set {1, 2, 3, 4}

// 交集
const intersection = new Set([...setA].filter(x => setB.has(x)));
console.log(intersection); // Set {2, 3}

// 差集
const difference = new Set([...setA].filter(x => !setB.has(x)));
console.log(difference); // Set {1}