LeetCode "4Sum"

Similar strategy could be applied to 4Sum, as 2sum to 3sum, but that'll be O(n^3). Instead of 3 + 1, we can also divide it into 2 + 2. So the problem becomes finding another pair - yes hashmap.

class Solution {
public:
    struct Rec
    {
        Rec() : i0(0), i1(0) {}
        Rec(int ri0, int ri1) : i0(ri0), i1(ri1) {}
        bool equals(Rec &r)
        {
            return i0 == r.i0 || i1 == r.i1 || i0 == r.i1 || i1 == r.i0;
        }
        int i0;
        int i1;
    };
    vector<vector<int>> fourSum(vector<int> &num, int target) {        
        vector<vector<int>> ret;
        if(num.size() < 4) return ret;

        //    Make pair
        size_t cnt = num.size();
        unordered_map<int, vector<Rec>> hm;
        for (size_t i = 0; i < cnt - 1; i ++)
        for (size_t j = i + 1; j < cnt; j++)
        {
            int sum2 = num[i] + num[j];
            if (hm.find(sum2) == hm.end())
            {
                vector<Rec> v; v.push_back(Rec(i, j));
                hm.insert(make_pair(sum2, v));
            }
            else
            {
                hm[sum2].push_back(Rec(i, j));
            }
        }
        
        //　　Match
        unordered_set<string> done;
        auto it = hm.begin();
        for (; it != hm.end(); it++)
        {
            int sum1 = it->first;
            
            auto it2 = hm.find(target - sum1);
            if (it2 != hm.end())
            {
                for (int i = 0; i < it2->second.size(); i ++)
                for (int j = 0; j < it->second.size(); j ++)
                {
                    Rec &r1 = it->second[j];
                    Rec &r2 = it2->second[i];
                    if (!r1.equals(r2))
                    {
                        vector<int> cv;
                        cv.push_back(num[r1.i0]); cv.push_back(num[r1.i1]);
                        cv.push_back(num[r2.i0]); cv.push_back(num[r2.i1]);
                        std::sort(cv.begin(), cv.end());

                        string str;
                        for (int i = 0; i < cv.size(); i++)
                        {
                            str += std::to_string(cv[i]);
                        }
                        if (done.find(str) == done.end())
                        {
                            ret.push_back(cv);
                            done.insert(str);
                        }
                    }
                }
            }
        }
        
        return ret;
    }
};