Answer 1

let arr0 = [1, 2, 3, '5', '你好啊', 7],
    arr1 = [2, 7, 9, 3, 'hello', '你好啊', 120];
    

const dorseyHandle = (arr0, arr1) => {

    let res = [],
        map = {},
        _arr1;

    //  确定哪个数组长度更长
    let _arr0 = arr0.length > arr1.length ? ( _arr1 = arr1, arr0 ) : ( _arr1 = arr0 , arr1 );    //  这里是arr0更长

    _arr0.map(( item, index ) => {

        map[item] = 'undefined' !== typeof map[item] ? map[item] + 1 : 1;
    });

    _arr1.forEach(item => {

        if('undefined' !== typeof map[item]){

            res.push(item);

            map[item] === 1 ? delete map[item] : map[item] -= 1;

        }
    });

    return res;
}

console.log(dorseyHandle(arr0, arr1));

另一种方式

    let arr0 = [1, 2, 3, '5', '你好啊', 7],
        arr1 = [2, 7, 9, 3, 'hello', '你好啊', 120];

    const dorseyHandle = ( arr0, arr1 ) => {


        return arr0.filter( item => {

            let index = arr1.indexOf(item);

            if(-1 !== index){

                arr1.splice(index, 1);
                return item;
            }
        });
    }

    console.log(dorseyHandle(arr0, arr1));

Answer 2

function reverse(num){ return (num + "").length ? (reverse((num + "").slice(1)) + (num + "")[0]) : "" }

Answer 3

处理数组

var a = [1, 2, 3, 5]
var b = [2, 4, 5, 1]
var c = [1, 3, 5]
var intersect
function fn(...arg){
  intersect =  arg.reduce((total,next)=>{return total.filter(item=>next.includes(item))})
}
fn(a,b,c)
console.log(intersect)
// [1,5]

处理数组和类数组（有iterable接口的数据结构）

var intersect
function fn2 (...arg){
    intersect = arg.reduce((total,next)=>{return [...total].filter(item=>new Set(next).has(item))})
}
var a = [1, 2, 3, 5]
var b = [2, 4, 5, 1]
var c = [1, 3, 5]
fn2(a,b,c)
console.log(intersect)
// [1,5]

a = new Set([1, 2, 3, 5])
b = new Set([2, 4, 5, 1])
c = new Set([1, 3, 5])
fn2(a,b,c)
console.log(intersect)
//[1,5]

Answer 4

var b = 10;
(function b(b){
console.log(b);
b = 20;
console.log(b);
})(b);