4Sum II

https://leetcode.com/submissions/detail/153740275/

class Solution {
public:
    int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {
        unordered_map<int,int> CD_freq;
        for (auto i : C)
            for (auto j : D)
                CD_freq[i+j]++;
        
        int res = 0;
        for (auto a : A)
            for (auto b : B) {
                auto it = CD_freq.find(-(a+b));
                if (it != CD_freq.end())
                    res += it->second;
            }
        return res;
    }
    int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {
        unordered_map<int,int> CD_freq;
        for (auto i : C)
            for (auto j : D)
                CD_freq[i+j]++;
        
        int res = 0;
        for (auto a : A)
            for (auto b : B)
                res += CD_freq[-(a + b)];　　// If CD_freq[-(a+b)] doesn't exist, it will be added! So map becomes much bigger!!!
        return res;
    }
    int fourSumCount_n3(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {
        unordered_map<int,int> D_freq;
        for (auto i : D)
            D_freq[i]++;
        
        int res = 0;
        for (int i = 0; i < A.size(); i++)
            for (int j = 0; j < B.size(); j++)
                for (int k = 0; k < C.size(); k++)
                    res += D_freq[-(A[i] + B[j] + C[k])];
        return res;
    }
    int fourSumCount_n4(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {
        int res = 0;
        for (int i = 0; i < A.size(); i++)
            for (int j = 0; j < B.size(); j++)
                for (int k = 0; k < C.size(); k++)
                    for (int l = 0; l < D.size(); l++)
                        if (A[i] + B[j] + C[k] + D[l] == 0)
                            res++;
        return res;
    }
};