454. 4Sum II -M

Given four lists A, B, C, D of integer values, compute how many tuples (i, j, k, l) there are such that A[i] + B[j] + C[k] + D[l] is zero.

To make problem a bit easier, all A, B, C, D have same length of N where 0 ≤ N ≤ 500. All integers are in the range of -228 to 228 - 1 and the result is guaranteed to be at most 231 - 1.

Example:

Input:
A = [ 1, 2]
B = [-2,-1]
C = [-1, 2]
D = [ 0, 2]

Output:
2

Explanation:
The two tuples are:
1. (0, 0, 0, 1) -> A[0] + B[0] + C[0] + D[1] = 1 + (-2) + (-1) + 2 = 0
2. (1, 1, 0, 0) -> A[1] + B[1] + C[0] + D[0] = 2 + (-1) + (-1) + 0 = 0

A:

-----------------用map --------------------

class Solution {
public:
    int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {
        unordered_map<int,int> AB;
        for(auto a:A)
            for(auto b:B)
                AB[a+b]++;
        
        unordered_map<int,int> CD;
        for(auto c: C)
            for(auto d: D)
                CD[c+d]++;
        int res=0;
        for(auto const& v:AB)
            res += v.second * CD[-v.first];
        return res;
    }
};

--------------改进一点儿------------  CD 那里不保存了。而只是-------

class Solution {
public:
    int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {
        unordered_map<int,int> AB;
        for(auto a:A)
            for(auto b:B)
                AB[a+b]++;
        
        int res = 0;
        for(auto c: C)
            for(auto d: D)
                res += AB[-c -d];
        return res;
    }
};