题目描述：
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分别从四个数组中取一个数，让它们的和等于0，求共有多少种组合。先用哈希表存储C和D中两数之和以及对应的次数，之后再求A和B中两数之和的相反数在哈希表中对应的次数，这样确定的四个数之和为0，将这个次数加入到最终结果里。
class Solution {
public:
int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {
int n=A.size();
unordered_map<int,int> hash;
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
hash[C[i]+D[j]]++;
int count=0;
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
count+=hash[-A[i]-B[j]];
return count;
}
};