腾讯&leetcode：给定两个数组，编写一个函数来计算它们的交集

[sisterAn]

给定两个数组，编写一个函数来计算它们的交集。

示例 1:

输入: nums1 = [1,2,2,1], nums2 = [2,2]
输出: [2]

示例 2:

输入: nums1 = [4,9,5], nums2 = [9,4,9,8,4]
输出: [9,4]

说明:

输出结果中的每个元素一定是唯一的。
我们可以不考虑输出结果的顺序。

附leetcode地址：leetcode

[finghtingjie]

const fn = (n1,n2)=>[...new Set(n1.filter(i => n2.includes(i)))]; console.log(fn([1,2,2,1],[2,2]))

[ai977313677]

用Map是O(n)的时间

var intersection = function(nums1, nums2) {
    const map = {}, ans = [];
    nums1.forEach(element => {
        map[element] = true;
    });
    nums2.forEach(element => {
        if (map[element]) {
            ans.push(element);
        }
    });
    return ans;
}

[zhdxmw]

var intersection = function(nums1, nums2) {
    let map1 = new Set(nums1);
    let map2 = new Set(nums2);
    let res = []
    map1.forEach((item) => {
        if(map2.has(item)){
             res.push(item)
        }
       
    })
    return res
};

[liwuhou]

const getIntersection = (nums1, nums2) => {
    return Array.from(new Set(nums1.filter(item => !!~nums2.indexOf(item))));
}

[Aiyoweioo]

/*找交集*/
function intersection(arr1, arr2){
 var hash = {}, _intersection = []
  for(let ele of arr1){
    if(!hash.hasOwnProperty(ele)){
      hash[ele] = true
    }
  }
  for(let ele of arr2){
    if(hash[ele]){
      var flag = false // NaN标志 
      if(_intersection.indexOf(ele) === -1){
        if(ele !== ele){
          if(!flag){
            flag = true
            _intersection.push(ele)
          }
        }else{
          _intersection.push(ele)
        }
      }  
    }
  }
  return _intersection
}
console.log(intersection([1,2,2,1], [2,2])) // [2]
console.log(intersection([4,9,5], [9,4,9,8,4])) // [9,4]
console.log(intersection([4,9,5,NaN], [9,4,9,8,4,NaN])) // [9,4,NaN]

/*找并集*/
function union(arr1,arr2){
  return Array.from(new Set([...arr1, ...arr2]))
}
console.log(union([1,2,2,1], [2,2])) // [1,2]
console.log(union([4,9,5], [9,4,9,8,4])) // [4,9,5,8]
console.log(union([4,9,5,NaN], [9,4,9,8,4,NaN])) // [4,9,5,NaN,8]

/*找差集*/
function difference(arr1, arr2){
  var set1 = new Set(arr1)
  var set2 = new Set(arr2)
  for(let ele of set1){
    if(set2.has(ele)){
      set2.delete(ele)
    }
  }
  return Array.from(set2)
}
console.log(difference([1,2,2,1], [2,2])) // []
console.log(difference([4,9,5], [9,4,9,8,4])) // [8]
console.log(difference([4,9,5,NaN], [9,4,9,8,4,NaN])) // [8]

[000Sonic]

var fn = (nums1, nums2) => {
    var small = nums1, big = nums2;
    if (nums1.length > nums2.length) {
        small = nums2;
        big = nums1;
    }

    return Array.from(new Set(small.filter(t => big.indexOf(t) > -1)));
};

console.log(fn([1, 2, 2, 1], [2, 2])) // [2]
console.log(fn([4,9,5], [9,4,9,8,4])) // [4,9]

[liwuhou]

var fn = (nums1, nums2) => {
    var small = nums1, big = nums2;
    if (nums1.length > nums2.length) {
        small = nums2;
        big = nums1;
    }

    return small.filter(t => big.indexOf(t) > -1);
};

fn([1, 2, 3, 4], [2, 6])

这个好像不满足

fn([1,2,3,4], [2,2]) // [2,2]

[000Sonic]

var fn = (nums1, nums2) => {
    var small = nums1, big = nums2;
    if (nums1.length > nums2.length) {
        small = nums2;
        big = nums1;
    }

    return small.filter(t => big.indexOf(t) > -1);
};

fn([1, 2, 3, 4], [2, 6])

这个好像不满足

fn([1,2,3,4], [2,2]) // [2,2]

改了，没注意去重

[uuRRx0]

function getIntersection(arr1, arr2) {
        var arr = [];
        arr1.forEach(function(elem) {
            arr2.includes(elem) && !arr.includes(elem) && arr.push(elem);
        })
        return arr;
    }

[sisterAn]

解题思路：

• filter 过滤
• Set 去重

代码实现：

const intersection = function(nums1, nums2) {
    return [...new Set(nums1.filter((item)=>nums2.includes(item)))]
};

leetcode

[hh-cheng]

function intersection(arr1, arr2) {
    if( arr1 instanceof Array && arr2 instanceof Array ){
        return Array.from(new Set(arr1.filter(item=>new Set(arr2).has(item))));
    }
    throw 'What I need are two arrays';
}

[JerryWen1994]

@WeCheung 你这filter里面，每一次都要new Set一遍，虽然代码简写了，但浪费了很多资源

[hh-cheng]

@597796340 多谢提醒

function intersection(arr1, arr2) {
    if( arr1 instanceof Array && arr2 instanceof Array ){
        const set = new Set(arr2);
        return Array.from(new Set(arr1.filter(item=>set.has(item))));
    }
    throw 'What I need are two arrays';
}

[lyllovelemon]

function combineArr(arr1,arr2) {
  let res=arr1.filter((item)=> arr2.includes(item))
    return new Set(res)
}