leetcode 454. 4Sum II 使用Map降低运算复杂的

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

题意很几简单，应该使用map来做标记来降低复杂度，嗯嗯，很不错的题，直接遍历肯定超时。

建议和这一道题leetcode 18. 4Sum KSum的解决办法 一起学习

代码如下：

#include <iostream>#include <vector>#include <map>#include <set>#include <queue>#include <stack>#include <string>#include <climits>#include <algorithm>#include <sstream>#include <functional>#include <bitset>#include <cmath>using namespace std;class Solution {public:    int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D)     {        int target = 0;        map<int, int> abSum;        for (int a : A)        {            for (int b : B)            {                abSum[a + b]++;            }        }        int count = 0;        for (int c : C)        {            for (int d : D)            {                if (abSum.find(target - c - d) != abSum.end())                    count += abSum[target - c - d];            }        }        return count;    }};