454. 4Sum II 难度：medium

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本文介绍了一种解决四数求和为零的问题的方法，利用两个哈希表降低时间复杂度至O(N^2)，并提供了C++实现代码。

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题目：

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

思路：

用空间换时间，使用两个map，时间复杂度是O（N^2）。

程序：

class Solution {
public:
    int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {
        int n=A.size();
        if(0==n)
            return 0;
        unordered_map<int,int>map1;
        unordered_map<int,int>map2;
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++)
            {
                int add=A[i]+B[j];
                map1[add]++;
            }
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++)
            {
                int add=C[i]+D[j];
                map2[add]++;
            }
        int count=0;
        for(auto it=map1.begin();it!=map1.end();it++)
        {
            int target=it->first;
            if(map2.find(-target)!=map2.end())
                count=count+map1[target]*map2[-target];
        }
        return count;
    }
};