第 142 题：（算法题）求多个数组之间的交集 #293

igoryaodev · 2019-10-16T07:55:08Z

算法

igoryaodev · 2019-10-16T07:57:01Z

142

1042478910 · 2019-10-17T01:56:42Z

let a = new Set([1, 2, 3]);
let b = new Set([4, 3, 2]);
let intersect = new Set([...a].filter(x => b.has(x)));
// set {2, 3}

Kisnnnnn · 2019-10-17T01:58:14Z

let setData1 = [1, 2, 3, 4],
  setData2 = [2, 3, 4, 9, 19, 91];

function mixedArr(...args) {
  let res = new Set(args[0]);
  if (args.length < 2) {
    return args[0];
  }
  for (let i = 1; i < args.length; i++) {
    const item = args[i];
    if (Object.prototype.toString.call(item).slice(8, -1) === 'Array') {
      let _arr = item.filter(e => res.has(e));
      if (_arr != [...res]) {
        res = new Set(_arr);
      }
    }
  }

  return [...res];
}

mixedArr(setData1, setData2, [2, 3]);

littleLane · 2019-10-17T02:00:34Z

function intersect(...args) {
  if (args.length === 0) {
    return []
  }

  if (args.length === 1) {
    return args[0]
  }

  return args.reduce((result, arg) => {
    return result.filter(item => arg.includes(item))
  })
}

zjtt · 2019-10-17T02:08:13Z

let getMix = arr => {
	return arr.reduce((total, item, index) => {
        if (index === 0) {
            return [...total, ...item]
        }
        else {
            return total.filter(v => item.includes(v))
        }
    }, [])
}
getMix([[1],[1,2],[1,3]]) //[1]

zxc6881553 · 2019-10-17T02:18:02Z

songkangle0826 · 2019-10-17T02:36:01Z

function intersection(...args){
if(args.length == 0) return [];
if(args.length == 1) return args[0];
let arr = [];
args.forEach((item)=>{
arr = arr.concat([...item]);
})
// 并集
// return new Set(arr);

// 交集
let obj = {};
let result = [];
for(let i in arr){
	if(obj[arr[i]]){
		result.push(arr[i])
	}else{
		obj[arr[i]] = true;
	}
}
return result;

}
intersection([1,2,3],[2,3]) // [2,3]

nowherebutup · 2019-10-17T02:49:02Z

const arr1 = [1, 2, 3];
const arr2 = [3, 4, 5];
const arr3 = [3, 6, 7];
const handle = (...arr) => {
  return arr.reduce((rst, ele, i) => {
    return rst.filter(item => ele.includes(item));
  });
}

handle(arr1, arr2, arr3);

DarthVaderrr · 2019-10-18T05:11:10Z

输入的数组有可能出现重复数据,需要多一步判断;如果输入的是集合,就不用这么复杂了

function intersectN(...arrs) {
    if (arrs.length === 1) return arrs[0];
    if (arrs.length === 2) return intersect2(...arrs);//如果只有两个 无需reduce 直接求
    return arrs.reduce((a, b) => b = intersect2(a, b))
}

function intersect2(nums1, nums2) {
    //求两个数组的交叉数组
    let res = [];
    let obj = toMap(nums1);
    for (let i of nums2) {
        if (obj[i]) {
            res.push(i)
            obj[i]--
        }
    }
    return res;
}

function toMap(arr) {
    //辅助函数  用于将数组转成map 键值分别是元素和其数量
    let obj = {}
    for (let i of arr) {
        if (obj[i]) obj[i]++
        else obj[i] = 1
    }
    return obj
}

let arr1 = [1, 2, 3, 4, 5, 6, 7, 1, 5]
let arr2 = [3, 5, 4, 1, 5]
let arr3 = [5, 7, 1, 3, 1, 5]

console.log(intersectN(arr1, arr2, arr3)) //[ 5, 1, 3, 5 ]

EnergySUD · 2019-10-18T06:23:05Z

function intersection(){
	let min_arr=arguments[0],intersect=[];
	for (let i=0;i<arguments.length;i++) {
		if(min_arr.length > arguments[i].length){ min_arr = arguments[i];}
	}
	for(let i=0;i<min_arr.length;i++){
		let flag = true;
		for (let j=0;j<arguments.length;j++) {
			if(!arguments[j].includes(min_arr[i])){
				flag = false;break;
			}
		}
		if(flag){ 
			if(!intersect.includes(min_arr[i])){ 
				intersect.push(min_arr[i]) 
			}
		}
	}
	return intersect
}
let arr = [1,2,3,4,5,6,7,8,9,10,10],arr1=[5,4,2,9,3,7,21],arr2=[9,3,5,4,11,3];
console.log(intersection(arr,arr1,arr2)) // [9, 3, 5, 4]

guestccc · 2019-10-18T17:48:06Z

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]

Minsaint · 2019-10-19T15:16:39Z

function getSameItem(arr, ...rest) {
	return arr.filter(item => rest.every(restArr => restArr.indexOf(item) > -1))
}

igoryaodev · 2019-10-20T09:25:56Z

const p1 = [
{
id: 123
},
{
id: 124
},
{
id: 125
},
]
const p2 = [
{
id: 124
},
{
id: 125
},
{
id: 125
},
{
id: 126
},
]
const p3 = [
{
id: 125
},
{
id: 128
},
{
id: 128
},
]
let a = intersections([p1, p2, p3])
console.log(a)

function intersections(options) {
  if (!Array.isArray(options) || !options.length) return;
  let arrLength = options.length;
  let result = [];
  let cacheFirstResult = [];
  let arr = [];
  let l = 0;
  let obj = {};
  let obj1 = {}
  
  function initState(arr) {
    let l = arr.length;
    while(l) {
      l--;
      obj1[arr[l].id] = true;
    }
  }
  initState(options[0])

  while (arrLength--) {
    arr = options[arrLength];
    l = arr.length;
    while (l) {
      l--;
      if (obj1[arr[l].id]) {
        obj[arr[l].id] = true;
        if (arrLength === options.length - 1)
          cacheFirstResult.push(arr[l])
      }
    }
    obj1 = {...obj}
    obj = {}

  }
  return cacheFirstResult.filter(v => {
    if (Object.keys(obj1).indexOf(v.id.toString()) !== -1) return v;
  })
}

snoweini · 2019-10-24T06:32:14Z

function setArr(...arg){
if(arg.length===0){
return false
}
if(arg.length===1){
return arg[0]
}
if(arg.length>1){
for(var i = 1; i<arg.length; i++){
arg[i] = selectArr(arg[i-1],arg[i])
}

}
return arg[arg.length-1];
}
console.log(setArr(arr2,arr3))

qq1037305420 · 2019-10-24T15:52:34Z

如果是对象数组上面是不是好多都不管用了

muzishuiji · 2019-10-27T11:18:17Z

function getSamePart() {
    let samePart = [], countMap = new Map();
    let args = [...arguments], len = args.length;
    args.reduce((arr, cur) => {
        return Array.isArray(cur) ? arr.concat(cur) : arr;
    }, []).forEach(item => {
        countMap.has(item) ? 
                 countMap.set(item, countMap.get(item) + 1) : countMap.set(item, 1)
    })
    for(let item of countMap) {
        if(item[1] >= len) {
            samePart.push(item[0])
        }
    }
    return samePart;
}
getSamePart(['1', '2', 1,2,3], ['1','2',1,2], ['1','2',3,4,5]);    // ['1', '2']

azl397985856 · 2019-10-27T11:57:40Z

Clarify

Since the description kind of loose, So Let's make it clear.

Suppose that, all the list are unique , eg: [1,2,2,3] is not allow.Otherwise the solution could be quiet ugly.

Concise Solution - `reduce`

initially, we can set the answer to be the very first item of the list.

eg: [1,2,3], [1,2,3,5], [1] , we take [1,2,3] as the initial value(the answer can't be greater than it). then we keep filtering repeatedly

If U'r not familar with reduce, PLZ check the MDN Doc

const intersection = (...list) => list.reduce((acc, cur) => acc.filter(ele => cur.includes(ele)))

// test
intersection([1,2,3], [1,2,3,5], [1]) // [1]

More efficient and Easy to understand

track the repeating count, store into Map as [k=item, v=count]
finally, remove all the items which repeating counts not equals to the size of the list(WHY?)

function intersection(...list) {
  const mapper = new Map();
  const res = [];
  for (ele of list) {
    for (n of ele) {
      if (mapper.get(n) === void 0) {
        mapper.set(n, 1);
      } else {
        mapper.set(n, mapper.get(n) + 1);
      }
    }
  }

  for ([k, cnt] of mapper.entries()) {
    if (cnt === list.length) res.push(k);
  }

  return res;
}

// test
intersection([1,2,3], [1,2,3,5], [1]) // [1]

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anjingdeyikexin · 2019-10-30T09:58:27Z

let a = [1, 2, 3];
let b = [4, 3, 2];
let result=a.filter((e)=>{
return b.includes(e)
})
console.log(result) //[2,3]

Fzzzzy · 2019-10-31T08:16:05Z

function findJoin(...rest) {
    if (rest.length === 0) return [];

    return Array.from(rest.reduce((result, arr) => {
        return new Set(result.filter(item => arr.includes(item)))
    }))
}

NANAYWY · 2019-11-01T02:01:08Z

function intersect(...arg) {
return arg.reduce((com, next) => com.filter(item => next.includes(item)));
}

fengshenhai-0727 · 2019-11-03T10:39:32Z

function inersection(a,b){
    if(!a.length || !b.length){
        return [];
    }
    const map = {};
    const result = [];
    for(let i=0; i<a.length; i++){
        map[a[i]] = map[a[i]] ? ++map[a[i]] : 1;
    }
    for(let i=0; i<b.length; i++){
        if(map[b[i]]){
            result.push(b[i]);
            map[b[i]]--;
        }
    }
    return result
}

function inersections(){
    if(arguments.length<2){
        return [];
    }
    const args = Array.prototype.slice.call(arguments);
    const result = [];
    let temp = inersection(args[0],args[1]);
    let i = 2;
    while(i<args.length){
        temp=inersection(temp,args[i])
        i++;
    }
    return temp;
}
inersections([1,2,2,1],[2,2],[2,2,4],[2]);//2

spongege · 2019-11-04T06:54:05Z

  const a = [1, 2, 2, 3]
  const b = [2, 2, 3, 4, 5]
  const c = [2, 2, 3, 6, 7, 2]

  const intersect = (...args) =>
    args.reduce(
      (acc, cur) => acc.filter(val => cur.includes(val)),
      args[0] || []
    )
  console.log(intersect(a, b, c))

lx812833 · 2019-11-05T03:53:08Z

const arr1 = [1, 2, 3];
const arr2 = [3, 4, 5];
const arr3 = [3, 6, 7];
const handle = (...arr) => {
  return arr.reduce((rst, ele, i) => {
    return rst.filter(item => ele.includes(item));
  });
}

handle(arr1, arr2, arr3);

在返回结果那里应该做一个去重

xiaowuge007 · 2019-12-05T07:11:10Z

let arr1 = [1,2,3,4,5];
        let arr2 = [2,4,6,8,7];
        let arr3 = [3,2,5,6,9,10,2];
        let arr4 = [10,23,4,56,1,2];
        let arr5 = [1,4,8,70,2];
        function fn(...arg) {
            let a = new Set(arg[0]);
            for(let i = 1;i<arg.length;i++){
                let arr = Array.from(a)
                for(let k = 0;k<arr.length;k++){
                    if(arg[i].indexOf(arr[k]) === -1){
                        a.delete(arr[k])
                    }
                }
            }
            return Array.from(a);
        }
        console.log(fn(arr1,arr2,arr3,arr4,arr5))

coveyz · 2019-12-09T09:33:14Z

给定两个数组，编写一个函数来计算它们的交集。

const intersect3 = (nums1,nums2) => { 
  let result = []
  for (const num of nums1) {
    let idx = nums2.indexOf(num)
    if (idx > -1) {
      result.push(num)
      nums2.splice(idx,1)
    }
  }
  return result
}

console.log(intersect3([4,9,5],[9,4,9,8,4]));

purplepeng · 2019-12-31T16:09:02Z

function intersect(...arrays) {
  let set = new Set(arrays[0])   
  for(let i=1; i< arrays.length; i++) {
    let tmpSet = new Set(arrays[i])
    set = new Set([...set].filter(x => tmpSet.has(x)))    
  }
  return [...set]
}

var items = intersect([1, 2, 3, 4], [2, 3, 4, 5], [3, 4, 5, 6])
console.log(items)

bbrucechen · 2020-01-03T03:13:29Z

function intersection(args1,...args) {
  args1 = [...new Set(args1)]
  const result = []
  for(let i = 0;i < args1.length;i ++) {
    let flag = true
    args.forEach(value => {
      if(!value.includes(args1[i])) {
        flag = false
      }
    })
    if(flag) {
      result.push(args1[i])
    }
  }
  return result
}

console.log(intersection([1,1,1,2,2,3,4,5,5],[1,2,3],[2,3,4],[6,9,2])) // [2]

goodmanforweb · 2020-01-10T15:16:52Z

Clarify

Since the description kind of loose, So Let's make it clear.

Suppose that, all the list are unique , eg: [1,2,2,3] is not allow.Otherwise the solution could be quiet ugly.

Concise Solution - reduce

initially, we can set the answer to be the very first item of the list.

eg: [1,2,3], [1,2,3,5], [1] , we take [1,2,3] as the initial value(the answer can't be greater than it). then we keep filtering repeatedly

If U'r not familar with reduce, PLZ check the MDN Doc
const intersection = (...list) => list.reduce((acc, cur) => acc.filter(ele => cur.includes(ele)))

// test
intersection([1,2,3], [1,2,3,5], [1]) // [1]
More efficient and Easy to understand

track the repeating count, store into Map as [k=item, v=count]

finally, remove all the items which repeating counts not equals to the size of the list(WHY?)
function intersection(...list) {
  const mapper = new Map();
  const res = [];
  for (ele of list) {
    for (n of ele) {
      if (mapper.get(n) === void 0) {
        mapper.set(n, 1);
      } else {
        mapper.set(n, mapper.get(n) + 1);
      }
    }
  }

  for ([k, cnt] of mapper.entries()) {
    if (cnt === list.length) res.push(k);
  }

  return res;
}

// test
intersection([1,2,3], [1,2,3,5], [1]) // [1]
Thanks for your reading, if like it, Give Me a thumb up 👍 and follow me on GitHub repo - frontend-interview

your code has bug,see example
intersection([1,2,1],[3,4])
// [1]

azl397985856 · 2020-01-10T17:36:57Z

Clarify

Since the description kind of loose, So Let's make it clear.
Suppose that, all the list are unique , eg: [1,2,2,3] is not allow.Otherwise the solution could be quiet ugly.

Concise Solution - reduce

initially, we can set the answer to be the very first item of the list.
eg: [1,2,3], [1,2,3,5], [1] , we take [1,2,3] as the initial value(the answer can't be greater than it). then we keep filtering repeatedly

If U'r not familar with reduce, PLZ check the MDN Doc
const intersection = (...list) => list.reduce((acc, cur) => acc.filter(ele => cur.includes(ele)))

// test
intersection([1,2,3], [1,2,3,5], [1]) // [1]
More efficient and Easy to understand

track the repeating count, store into Map as [k=item, v=count]

finally, remove all the items which repeating counts not equals to the size of the list(WHY?)
function intersection(...list) {
  const mapper = new Map();
  const res = [];
  for (ele of list) {
    for (n of ele) {
      if (mapper.get(n) === void 0) {
        mapper.set(n, 1);
      } else {
        mapper.set(n, mapper.get(n) + 1);
      }
    }
  }

  for ([k, cnt] of mapper.entries()) {
    if (cnt === list.length) res.push(k);
  }

  return res;
}

// test
intersection([1,2,3], [1,2,3,5], [1]) // [1]
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your code has bug,see example
intersection([1,2,1],[3,4])
// [1]

认真看我的前提，谢谢

wuaixiaoyao · 2020-03-12T04:35:11Z

function getIntersection (...args) {
  //判读是一个还是多个数组
  if (args.length > 1) {
    return args.reduce((pre, cur)=> {
       let preResult = [...new Set(pre)].filter(item => cur.includes(item) );
       //用数组的 includes 或者 set 的has方法验证
       //let preResult = [...new Set(pre)].filter(item => new Set(cur).has(item) );
       return preResult 
    })
  } else {  
    return args[0]
  }

}
console.log(getIntersection([1,2,2,3], [2,2,3,4,5],[3,4,5])
)

jojomango · 2020-03-24T02:07:23Z

const a = [1,2,3,1,4,5], b = [3,3,2,5], c = [1,1,1,3,2];
const uniq = arr => [...new Set(arr)];
const mix = (arg1,...arrs) => {
  const uniqArr = uniq(arg1);
  return arrs.reduce((result, arr) => {
    return result.filter(ele => arr.includes(ele));
  }, uniqArr);
};

console.log(mix(a,b,c)); // [2,3]

fzhange · 2020-05-10T13:49:04Z

function fun(...arg){
  let myArr = arg.map(item=>[...new Set(item)]
  )
  let myData = myArr.reduce((pre,curr)=>{
    return pre.map((item)=>{
      if(curr.indexOf(item) != -1) return item;
    }).filter((item)=>!!item)
  })
  return myData;
}

console.log('fun([1,2,2,3,4,4],[2,1,2,4],[2,1])  : ', fun([1,2,2,3,4,4],[2,1,2,4],[2,1])  ); //[1,2]

meta07 · 2020-05-15T02:58:07Z

参考@fengshenhai-0727大兄弟的，即使数组有重复的交集项也没问题。

function getTwoArrsCommon(arr1, arr2) {
    if(!arr1.length || !arr2.length) {
        return [];
    }
    const res = [];
    const map = {};
    for(let k of arr1) {
        if(map[k]) {
            map[k]++;
        } else {
            map[k] = 1;
        }
    }
    for(let k of arr2) {
        if(map[k] > 0) {
            res.push(k);
            map[k]--;
         }
    }
    return res;
}
function getManyArrsCommon() {
    if(arguments.length < 2) {
        return [];
    }
    const args = Array.from(arguments);
    let res = getTwoArrsCommon(args[0], args[1]);
    let i = 2;
    while(i < args.length) {
        res = getTwoArrsCommon(res, args[i]);
        i++;
    }
   return res;
}
getManyArrsCommon([1,2,2,1],[2,2],[2,2,4],[2]); // [2]
getManyArrsCommon([1,1,2,3], [1,1,1,2,3,5,5], [1,1]); // [1, 1]
getManyArrsCommon([1,1,3,5,5], [2, 4, 5], [3, 5, 5]); // [5]

cutie6 · 2020-06-13T00:16:36Z

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]

如果第一个数组中有重复的两个1呢，这样算出来，交集中也会有两个1吧

cutie6 · 2020-06-13T04:00:04Z

Clarify

Since the description kind of loose, So Let's make it clear.

Suppose that, all the list are unique , eg: [1,2,2,3] is not allow.Otherwise the solution could be quiet ugly.

Concise Solution - reduce

initially, we can set the answer to be the very first item of the list.

eg: [1,2,3], [1,2,3,5], [1] , we take [1,2,3] as the initial value(the answer can't be greater than it). then we keep filtering repeatedly

If U'r not familar with reduce, PLZ check the MDN Doc
const intersection = (...list) => list.reduce((acc, cur) => acc.filter(ele => cur.includes(ele)))

// test
intersection([1,2,3], [1,2,3,5], [1]) // [1]
More efficient and Easy to understand

track the repeating count, store into Map as [k=item, v=count]

finally, remove all the items which repeating counts not equals to the size of the list(WHY?)
function intersection(...list) {
  const mapper = new Map();
  const res = [];
  for (ele of list) {
    for (n of ele) {
      if (mapper.get(n) === void 0) {
        mapper.set(n, 1);
      } else {
        mapper.set(n, mapper.get(n) + 1);
      }
    }
  }

  for ([k, cnt] of mapper.entries()) {
    if (cnt === list.length) res.push(k);
  }

  return res;
}

// test
intersection([1,2,3], [1,2,3,5], [1]) // [1]
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中国人为啥要用英文

cutie6 · 2020-06-13T06:03:43Z

function intersection(...args){
if(args.length == 0) return [];
if(args.length == 1) return args[0];
let arr = [];
args.forEach((item)=>{
arr = arr.concat([...item]);
})
// 并集
// return new Set(arr);
// 交集
let obj = {};
let result = [];
for(let i in arr){
	if(obj[arr[i]]){
		result.push(arr[i])
	}else{
		obj[arr[i]] = true;
	}
}
return result;
}
intersection([1,2,3],[2,3]) // [2,3]

这种算法，[1,1,1,] 和 [] 算出来岂不是 [1,1,1]

cutie6 · 2020-06-13T06:15:52Z

function intersection(){
	let min_arr=arguments[0],intersect=[];
	for (let i=0;i<arguments.length;i++) {
		if(min_arr.length > arguments[i].length){ min_arr = arguments[i];}
	}
	for(let i=0;i<min_arr.length;i++){
		let flag = true;
		for (let j=0;j<arguments.length;j++) {
			if(!arguments[j].includes(min_arr[i])){
				flag = false;break;
			}
		}
		if(flag){ 
			if(!intersect.includes(min_arr[i])){ 
				intersect.push(min_arr[i]) 
			}
		}
	}
	return intersect
}
let arr = [1,2,3,4,5,6,7,8,9,10,10],arr1=[5,4,2,9,3,7,21],arr2=[9,3,5,4,11,3];
console.log(intersection(arr,arr1,arr2)) // [9, 3, 5, 4]

这样算的话，那[1,1,1,2] 和 [1,1,1,] 的交集只有1了

NathanHan1 · 2020-06-22T09:12:30Z

function intersection(...arr) {
        // 注意 交集是集合概念，集合是不存在重复元素的
        const map = {}
        const len = arr.length
        const result = []
        for(let i = 0; i < len; i++) {
          const subArr = arr[i]
          for (let j = 0 ; j< subArr.length; j++) {
            const v = subArr[j]
            if(!map[v]) {
              map[v] = 1
            } else {
              map[v] += 1
            }
          }
        }
        for(let key in map) {
          if(map[key] === len) {
            result.push(~~key)
          }
        }
        return result
      }

dice246 · 2020-06-26T01:25:51Z

// 求多个数组之间的交集
function intersection(arr1, arr2) {
  let s1 = new Set([...arr1]);
  let s2 = new Set([...arr2]);
  let result = []

  for (let item of s1.values()) {
    if (s2.has(item)) {
      result.push(item)
    }
  }

  return result;
}

const arr1 = [1,2,3,4,5];
const arr2 = [3,6,7];
const result = intersection(arr1, arr2);
console.log(result);

ustchcl · 2020-07-07T15:10:01Z

// 不使用set

function intersect<A>(arr1: Array<A>, arr2: Array<A>): Array<A> {
  return arr2.filter(x => arr1.indexOf(x) !== -1)
}

// 递归
function intersectAll<A>(arr: Array<Array<A>>): Array<A> {
  if (arr.length === 0) return []
  else if (arr.length === 1) return arr[0]
  else {
    const sp = Math.floor(arr.length / 2)
    return intersect(intersectAll(arr.slice(0, sp)), intersectAll(arr.slice(sp)))
  }
}

// 迭代
function intersectAllReduce<A>(arr: Array<Array<A>>): Array<A> {
  if (arr.length === 0) return []
  else return arr.slice(1).reduce((pv, cv) => intersect(pv, cv), arr[0])
}

gitwillsky · 2020-08-12T12:52:46Z

function intersect(...arrays) {
  if (!arrays) {
    return [];
  }
  if (arrays.length === 1) {
    return arrays[0];
  }

  const m = new Map();

  for (let i = 0; i < arrays.length; i++) {
    for (let j = 0; j < arrays[i].length; j++) {
      const key = arrays[i][j];
      const value = i + 1;
      if (i === 0) {
        m.set(key, value);
        continue;
      }
      if (m.has(key) && m.get(key) < value) {
        m.set(key, value);
      }
    }
  }
  return Array.from(m.keys()).filter((key) => m.get(key) === arrays.length);
}

shizhenbin · 2020-08-30T07:58:14Z

function intersect(...args) {
  if (args.length === 0) {
    return []
  }

  if (args.length === 1) {
    return args[0]
  }

  return args.reduce((result, arg) => {
    return result.filter(item => arg.includes(item))
  })
}

console.log(intersect([1,1,1,1,1,2,9,63],[43,1,3,8],[3,3,3,,1,2,4,1,2])) //[1, 1, 1, 1, 1] 这个得到的不算是交集吧

yangchaojie456 · 2020-08-31T06:24:46Z

你们都好聪明，我只会笨方法

console.log(intersection([1, 1, 1, 1, 1, 2, 9, 63], [43, 1, 3, 8], [3, 3, 3, , 1, 2, 4, 1, 2]))

function intersection() {
    var map = new Map()
    for (var i = 0; i < arguments.length; i++) {
        var arr = [...new Set(arguments[i])]
        for (var j = 0; j < arr.length; j++) {
            if (!map.has(arr[j])) {
                map.set(arr[j], 1)
            } else {
                map.set(arr[j], map.get(arr[j]) + 1)
            }
        }
    }
    var collect = []
    for (const iterator of map) {
        if (iterator[1] == arguments.length) {
            collect.push(iterator[0])
        }
    }
    return collect
}

Godguns · 2020-11-20T08:23:18Z

`function choose(...arr) {
let temp = [...arr].flat().sort((a, b) => {
return a - b;
});

let ret = [];
for (let i = 0; i < temp.length; i++) {
if (temp[i] == temp[i + 1]) {
ret.push(temp[i]);
}
}
console.log(ret);
}
let a = [1, 3, 4, 5];
let b = [1, 2, 3, 4, 5];
choose(a, b);
`

pan-Z2l0aHVi · 2021-01-11T09:28:41Z

/**
 * 数组交集
 * @param  {...Array} arrays
 */
function arrayIntersection(...arrays) {
  if (arrays.length === 0) return []
  if (arrays.length === 1) return arrays[0]

  return arrays.reduce(
    (acc, cur) => (acc = acc.filter((accItem) => cur.some((curItem) => curItem === accItem)))
    // 或者 (acc, cur) => (acc = acc.filter((accItem) => cur.includes(accItem)))
  )
}

/**
 * test
 */
const res = arrayIntersection([1, 2, 3, 4, 5], [3, 4, 5, 6, 7], [4, 5, 6, 7, 8])
console.log('res: ', res) // [4, 5]

wenchao0809 · 2021-03-09T10:39:31Z

用分治代码比较简洁，没考虑性能啥的

function intersection(...args) {
  if (args.length === 1) return args[0]
  if (args.length == 2) return [...new Set(args[0].filter(item => args[1].indexOf(item) !== -1))]
  const mid = Math.floor(args.length / 2)
  const left = intersection.apply(null, args.slice(0, mid))
  const right = intersection.apply(null, args.slice(mid))
  return intersection.apply(null, [left, right])
}

qzruncode · 2021-04-13T13:45:30Z

思想来源：借鉴vue diff源码
function intersectionArr(...params) {
    let itemCountMap = [];
    params.forEach(arr => {
      arr.forEach(item => {
        itemCountMap[item] = (itemCountMap[item] ? itemCountMap[item] : 0) + 1;
      })
    })

    let itArr = [];
    itemCountMap.forEach((item, index) => {
      if(item == params.length) {
        itArr.push(index);
      }
    })
    return itArr;
  }
  intersectionArr([1,2,3,4], [2,3], [2,3,4], [1,2,3] );

poppydani · 2021-08-18T09:44:11Z

function intersect(...args) {
  const flattened = args.reduce((res, arr) => {
    return res.concat(arr);
  }, []);
  return [...new Set(flattened)].filter((v) =>
    args.every((arr) => arr.includes(v))
  );
}

miniflycn · 2022-04-16T13:01:13Z

function mix(...arrs) {
    function createMap(arr) {
        return arr.reduce((result, value) => {
            result[value] = result[value] || 0
            result[value]++
            return result
        }, {})
    }

    const mapList = arrs.map(createMap)
    const res = mapList.shift()

    mapList.reduce((result, map) => {
        for (let key in result) {
            if (map[key]) {
                result[key] = Math.min(map[key], result[key])
            } else {
                delete result[key]
            }
        }
        return result
    }, res)

    const result = []

    for (let key in res) {
        while (res[key]--) {
            result.push(key)
        }
    }

    return result
}

Dylan0916 · 2022-04-17T03:00:05Z

function foo(...args) {
  return args.slice(1).reduce(
    (acc, cur) => {
      return acc.filter((v) => new Set(cur).has(v));
    },
    [...new Set(args[0])]
  );
}

foo([1, 2, 3], [3, 1, 4, 5, 6, 7], [1, 4, 5, 3]); // [1,3]

Yangfan2016 · 2022-08-30T13:58:53Z

https://leetcode.cn/problems/intersection-of-multiple-arrays/submissions/

// 法一： reduce 结合 includes 去重

/**
 * @param {number[][]} nums
 * @return {number[]}
 */
var intersection = function (nums) {
    return nums.reduce((prev, cur) => prev.filter(item => cur.includes(item))).sort((a, b) => a - b);
};

// 法二： map 统计每个元素的个数，如果个数等于多数组的总数，则认为存在交集

/**
 * @param {number[][]} nums
 * @return {number[]}
 */
var intersection = function (nums) {
    const map = new Map;
    const res = [];
    for (let i = 0; i < nums.length; i++) {
        for (let j = 0; j < nums[i].length; j++) {
            const item = nums[i][j];
            map.set(item, (map.get(item) || 0) + 1);
        }
    }
    for (let [key, value] of map) {
        if (value === nums.length) {
            res.push(key);
        }
    }
    return res.sort((a, b) => a - b);
};

igoryaodev changed the title ~~算法题：求多个数组之间的交集【阿里】~~ 第 142 题：（算法题）求多个数组之间的交集【阿里】 Oct 16, 2019

yygmind changed the title ~~第 142 题：（算法题）求多个数组之间的交集【阿里】~~ 第 142 题：（算法题）求多个数组之间的交集 Oct 17, 2019

yygmind added the 阿里 label Oct 17, 2019

yygmind added the 编程题 label Dec 16, 2019

第 142 题：（算法题）求多个数组之间的交集 #293

第 142 题：（算法题）求多个数组之间的交集 #293

Comments

igoryaodev commented Oct 16, 2019

igoryaodev commented Oct 16, 2019

1042478910 commented Oct 17, 2019

Kisnnnnn commented Oct 17, 2019

littleLane commented Oct 17, 2019 • edited Loading

zjtt commented Oct 17, 2019

zxc6881553 commented Oct 17, 2019

songkangle0826 commented Oct 17, 2019

nowherebutup commented Oct 17, 2019

DarthVaderrr commented Oct 18, 2019

EnergySUD commented Oct 18, 2019

guestccc commented Oct 18, 2019

处理数组

处理数组和类数组（有iterable接口的数据结构）

Minsaint commented Oct 19, 2019

igoryaodev commented Oct 20, 2019

snoweini commented Oct 24, 2019 • edited Loading

qq1037305420 commented Oct 24, 2019

muzishuiji commented Oct 27, 2019 • edited Loading

azl397985856 commented Oct 27, 2019

Clarify

Concise Solution - reduce

More efficient and Easy to understand

anjingdeyikexin commented Oct 30, 2019

Fzzzzy commented Oct 31, 2019

NANAYWY commented Nov 1, 2019

fengshenhai-0727 commented Nov 3, 2019

spongege commented Nov 4, 2019

lx812833 commented Nov 5, 2019

xiaowuge007 commented Dec 5, 2019

coveyz commented Dec 9, 2019

purplepeng commented Dec 31, 2019 • edited Loading

bbrucechen commented Jan 3, 2020

goodmanforweb commented Jan 10, 2020

Clarify

Concise Solution - reduce

More efficient and Easy to understand

azl397985856 commented Jan 10, 2020

Clarify

Concise Solution - reduce

More efficient and Easy to understand

wuaixiaoyao commented Mar 12, 2020 • edited Loading

jojomango commented Mar 24, 2020 • edited Loading

fzhange commented May 10, 2020

meta07 commented May 15, 2020 • edited Loading

cutie6 commented Jun 13, 2020

处理数组

处理数组和类数组（有iterable接口的数据结构）

cutie6 commented Jun 13, 2020

Clarify

Concise Solution - reduce

More efficient and Easy to understand

cutie6 commented Jun 13, 2020

cutie6 commented Jun 13, 2020

NathanHan1 commented Jun 22, 2020

dice246 commented Jun 26, 2020

ustchcl commented Jul 7, 2020

gitwillsky commented Aug 12, 2020

shizhenbin commented Aug 30, 2020

yangchaojie456 commented Aug 31, 2020

Godguns commented Nov 20, 2020 • edited Loading

pan-Z2l0aHVi commented Jan 11, 2021

wenchao0809 commented Mar 9, 2021

用分治代码比较简洁， 没考虑性能啥的

qzruncode commented Apr 13, 2021

poppydani commented Aug 18, 2021

miniflycn commented Apr 16, 2022 • edited Loading

Dylan0916 commented Apr 17, 2022 • edited Loading

Yangfan2016 commented Aug 30, 2022 • edited Loading

littleLane commented Oct 17, 2019 •

edited

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snoweini commented Oct 24, 2019 •

edited

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muzishuiji commented Oct 27, 2019 •

edited

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Concise Solution - `reduce`

purplepeng commented Dec 31, 2019 •

edited

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Concise Solution - `reduce`

Concise Solution - `reduce`

wuaixiaoyao commented Mar 12, 2020 •

edited

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jojomango commented Mar 24, 2020 •

edited

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meta07 commented May 15, 2020 •

edited

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Concise Solution - `reduce`

Godguns commented Nov 20, 2020 •

edited

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用分治代码比较简洁，没考虑性能啥的

miniflycn commented Apr 16, 2022 •

edited

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Dylan0916 commented Apr 17, 2022 •

edited

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Yangfan2016 commented Aug 30, 2022 •

edited

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