LeetCode 4Sum II 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

---

本题和LeetCode 4Sum类似，只不过给定的是四个数组，问分别从四个数组中取一个数，加起来和为0的方案总数。 暴力方法是$$O(n^4)$$的，但是我们可以利用Hash方法将复杂度降为$$O(n^2)$$。分别对A[i]+B[j]和C[i]+D[j]出现的次数建立Hash，然后遍历其中一个Hash表，在另一个Hash表中查相反数出现的次数，然后乘起来。代码如下： [cpp] class Solution { public: int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) { int ans = 0, n = A.size(); unordered_map<int, int> hash1, hash2; for(int i = 0; i < n; ++i){ for(int j = 0; j < n; ++j){ ++hash1[A[i] + B[j]]; ++hash2[C[i] + D[j]]; } } for(unordered_map<int, int>::iterator it = hash1.begin(); it != hash1.end(); ++it){ ans += it->second * hash2[-it->first]; } return ans; } }; [/cpp] 本代码提交AC，用时909MS。 参考：http://www.cnblogs.com/grandyang/p/6073317.html]]>