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
这道题是之前那道4Sum的延伸，让我们在四个数组中各取一个数字，使其和为0。那么坠傻的方法就是遍历所有的情况，时间复杂度为O(n4)。但是我们想想既然Two Sum那道都能将时间复杂度缩小一倍，那么这道题我们使用哈希表是否也能将时间复杂度降到O(n2)呢？答案是肯定的，我们如果把A和B的两两之和都求出来，在哈希表中建立两数之和跟其出现次数之间的映射，那么我们再遍历C和D中任意两个数之和，我们只要看哈希表存不存在这两数之和的相反数就行了，参见代码如下：
解法一：
class Solution { public: int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) { int res = 0; unordered_map<int, int> m; for (int i = 0; i < A.size(); ++i) { for (int j = 0; j < B.size(); ++j) { ++m[A[i] + B[j]]; } } for (int i = 0; i < C.size(); ++i) { for (int j = 0; j < D.size(); ++j) { int target = -1 * (C[i] + D[j]); res += m[target]; } } return res; } };
这种方法用了两个哈希表分别记录AB和CB的两两之和出现次数，然后遍历其中一个哈希表，并在另一个哈希表中找和的相反数出现的次数，参见代码如下：
解法二：
class Solution { public: int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) { int res = 0, n = A.size(); unordered_map<int, int> m1, m2; for (int i = 0; i < n; ++i) { for (int j = 0; j < n; ++j) { ++m1[A[i] + B[j]]; ++m2[C[i] + D[j]]; } } for (auto a : m1) res += a.second * m2[-a.first]; return res; } };
本文转自博客园Grandyang的博客，原文链接：四数之和之二[LeetCode] 4Sum II ，如需转载请自行联系原博主。
