4sumⅡ

Given four lists A, B, C, D of integer values, compute how many tuples (i, j, k, l) there are such thatA[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

这道题最暴力直接的思路就是用四个循环，然后计算是否有和为0的一组数

但这样肯定会超时，所以我们可以把这四组数变为两个哈希表，然后用O（1）的算法找到和为0的那种情况，最终的时间复杂度O（n^2)

代码如下：

class Solution {public:    void sum(vector<int> &a,vector<int> &b,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]]++;        }    }    int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {        unordered_map<int,int> m1;        unordered_map<int,int> m2;        sum(A,B,m1);        sum(C,D,m2);        int ret=0;        for(auto i=m1.begin();i!=m1.end();i++)        {            auto j=m2.find(-i->first);            if(j!=m2.end())             {                 ret=ret+i->second*(j->second);             }        }        return ret;    }};