第 142 题：（算法题）求多个数组之间的交集

[igoryaodev]

算法

[igoryaodev]

142

[1042478910]

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]

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]

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]

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]

let array1 = [1,2,3,4,5,6];
let array2 = [2,3,4,5,6,8,9];
let array3 = [4,5,6,7,8,9]
console.log(getIntersection(array1, array2, array3));
function getIntersection(){
let result = [];
let obj = {};
let array = [].slice.apply(arguments)
for(let i = 0; i < array.length; i++) {
let arrayItem = array[i]
for(let n = 0; n < arrayItem.length; n++) {
if (obj[arrayItem[n]]) {
obj[arrayItem[n]] = obj[arrayItem[n]] + 1
} else {
obj[arrayItem[n]] = 1
}
}
};
for(let item in obj) {
if(obj[item] > 1) {
result.push(item)
}
};
return result;
};

[songkangle0826]

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]

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]

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]

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]

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]

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