最效率获得俩个数组的相同值

Question

let arr1=['a','b','c','d','e']let arr2=['c','d','g','h','a']找到arr1和arr2中相同的值，并在arr2中删除对应的值怎么写才更简单效率，我只能想到套个双层循环，就比如这样    arr1.forEach((val)=>{        arr2.forEach((v,i)=>{            val === v &&  arr2.splice(i,1)        })    })然后现在问题变了arr2=[{'id':'c'},{'id':'d'},{'id':'g'},{'id':'h'},{'id':'a'}]我的只要改一个地方就能继续运行arr1.forEach((val)=>{    arr2.forEach((v,i)=>{        val === v.id &&  arr2.splice(i,1)    })})你们代码需要怎么改才能实现原来的目的，改动最少最简单的，我认为是最合理的就采纳了

Answer 1

let arr1=['a','b','c','d','e']

let arr2=['c','d','g','h','a']

let res = []

var a = arr2.filter(v => {

    if (arr1.indexOf(v) >= 0) {

        res.push(v)

    }

    return arr1.indexOf(v) < 0

})

res // ["c", "d", "a"]

a // ["g", "h"]

Answer 2

array 转 dict

let arr1=['a','b','c','d','e']

let arr2=['c','d','g','h','a']

let obj = arr1.reduce((iter, val) => {

    iter[val] = true

    return iter

}, {})

let result = arr2.filter(v => !obj[v])

console.log(result)