LeetCode 454. 4Sum II (Medium)

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:2Explanation:The two tuples are:1. (0, 0, 0, 1) -> A[0] + B[0] + C[0] + D[1] = 1 + (-2) + (-1) + 2 = 02. (1, 1, 0, 0) -> A[1] + B[1] + C[0] + D[0] = 2 + (-1) + (-1) + 0 = 0

题目大意：给出四个大小相同的数组，找出有多少组数字（系数），使每个数组中任取一个元素相加等于0。

思路：将前两个数字的数字相加做一下哈希，得出任意前两个数组的数字和，再遍历后两个数组找有没有匹配的（哈希）。处理相同情况就把哈希的值+1。

c++代码：

class Solution {public:    int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {        map<int, int> hash;        int ans = 0;        for (auto i : A)        {            for (auto j : B)            {                hash[i + j]++;            }        }        for (auto i : C)        {            for (auto j : D)            {                if (hash.count(-(i + j)) != 0)                {                    ans += hash[-(i + j)];                }            }        }        return ans;    }};