Leetcode 350. Intersection of Two Arrays II

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Given two arrays, write a function to compute their intersection.

Example:
Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2, 2].

Hash table solution: Time: O(m + n) Space: O(m + n)

class Solution {public:    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {        unordered_map<int, int> dict;        vector<int> res;        for(int i = 0; i < (int)nums1.size(); i++) dict[nums1[i]]++;        for(int i = 0; i < (int)nums2.size(); i++)            if(--dict[nums2[i]] >= 0) res.push_back(nums2[i]);        return res;    }};

Hash table solution2: Time: O(m + n) Space: O(m)

class Solution {public:    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {        unordered_map<int, int> dict;        vector<int> res;        for(int i = 0; i < (int)nums1.size(); i++) dict[nums1[i]]++;        for(int i = 0; i < (int)nums2.size(); i++)        if(dict.find(nums2[i]) != dict.end() && --dict[nums2[i]] >= 0)              res.push_back(nums2[i]);        return res;    }};

Sort and two pointers Solution: Time: O(max(m, n) log(max(m, n))) Space: O(m + n)

class Solution {public:    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {        sort(nums1.begin(), nums1.end());        sort(nums2.begin(), nums2.end());        int n1 = (int)nums1.size(), n2 = (int)nums2.size();        int i1 = 0, i2 = 0;        vector<int> res;        while(i1 < n1 && i2 < n2){            if(nums1[i1] == nums2[i2]) {                res.push_back(nums1[i1]);                i1++;                i2++;            }            else if(nums1[i1] > nums2[i2]){                i2++;            }            else{                i1++;            }        }        return res;    }};

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