请问大家一个小算法问题

Question

var arr1 = ['27709', '29406', '29407', '29408', '29409', '29410', '29411', '29412', '29414', '29415']

var arr2 =  ['27709', '29406', '29409', '29410', '29411', '29412', '29414', '29415']

上面两个数组，arr1里有10个，arr2里有8个，请问如何找出包含arr2中的，多出来的两个值，谢谢大家抽空帮忙解答

Answer 1

是找出差值 不包含的吧

let newArr = arr1.filter(item=>!arr2.includes(item))

Answer 2

直接上菜

function subArr(a, b) {
    return a.filter(i => b.indexOf(i) == -1)
}
const arr1 = ['27709', '29406', '29407', '29408', '29409', '29410', '29411', '29412', '29414', '29415']
const arr2 = ['27709', '29406', '29409', '29410', '29411', '29412', '29414', '29415']
console.log(subArr(arr1, arr2))

Answer 3

function partition() {
    var intersection = arguments[0], complement = [];
    for (var i = 1; i < arguments.length; ++i) {
        var temporary = [];
        for (var j = 0; j < arguments[i].length; ++j) {
            var item = arguments[i][j];
            if (intersection.indexOf(item) >= 0) {
                temporary.push(item);
            } else if (complement.indexOf(item) < 0) {
                complement.push(item);
            }
        }
        intersection = temporary;
    }
    for (var i = arguments[0].length; i--;) {
        var item = arguments[0][i];
        if (intersection.indexOf(item) < 0 && complement.indexOf(item) < 0) {
            complement.unshift(item);
        }
    }
    return {
        intersection: intersection,
        complement: complement,
        union: intersection.concat(complement)
    };
}
var set = partition(arr1, arr2);
console.log(set.complement);        // 补集
console.log(set.intersection);      // 交集
console.log(set.union);             // 并集