js多个数组取交集(三个及以上)

// 原数组
const serveralArr = [
    [1,2,4,5,23,3,2,2,4,3,5,5], 
    [3,2,3,2,2,4,3,1,4,5,6], 
    [3,2,4,3,2,4,1,2,5],
    [3,2,4,5,5,4,3,1,2,2],
    [3,2,23,3,4,1,3,4,5,5,4,3,1,2,2],
    [3,2,4,1,2,5,5,4,3,1,2,2],
    [3,2,4,25,5,4,3,1,2,2],
]
// ES5 方法实现数学意义上的交集结果
const intersectNoRepeatFirst = serveralArr => {
    let minArr = serveralArr[0]
    serveralArr.forEach(item => {
        if(item.length < minArr.length){
            minArr = item
        }
    })
    const result = []
    minArr.forEach(item => {
        serveralArr.forEach(j => {
            if(j.includes(item) && !result.includes(item)){
                result.push(item)
            }
        })
    })
    return result
}
// ES6 方法实现数学意义上的交集结果
const intersectNoRepeatTwice = arrs => {
    return arrs.reduce(function(prev,cur){
        // return [...new Set(cur.filter((item)=>prev.includes(item)))]
        return Array.from(new Set(cur.filter((item)=>prev.includes(item))))
    })
}
// 输出数学意义上的交集结果
console.log(intersectNoRepeatFirst(serveralArr), intersectNoRepeatTwice(serveralArr)) // => (5) [3, 2, 4, 1, 5]; (5) [3, 2, 4, 5, 1]
// ES5 方法实现有重复元素的交集结果
const intersectRepeat = arr => {
    const result = []
    const tempArr = []
    arr.forEach(item => {
        const obj = {}
        item.forEach(j => {
            if(obj.hasOwnProperty(j)){
                obj[j] += 1
            }else{
                obj[j] = 1
            }
        })
        tempArr.push(obj)
    })
    let arrr = tempArr[0]
    tempArr.forEach(item => {
        if(item.length < arrr.length){
            arrr = item
        }
    })
    const newOjb = {}
    Object.keys(arrr).forEach(item => {
        newOjb[item] = Math.min.apply(Math, tempArr.map(function(o) {return o[item]}))
    })
    Object.keys(newOjb).forEach(item => {
        if(newOjb[item]){
            for(let i = 0; i < newOjb[item]; i++){
                result.push(Number(item))
            }
        }
    })
    return result
}
// 输出有重复元素的交集结果
console.log(intersectRepeat(serveralArr)) // => (9) [1, 2, 2, 2, 3, 3, 4, 4, 5]