454. 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

给定4个整型数组A、B、C、D，统计有多少个4元组（i,j,k,l），使它们满足A[i]+B[j]+C[k]+D[l]等于0。
为简化问题，假定这3个数组均含N个元素，且 0<=N<=500。各数组所含的元素均在(-2)^28到(2^28-1)之间。且结果不超过(2^31-1)。
分析：
这道题按我自己的想法是依次遍历各个数组，当然这是很蠢的，惭愧！最终提交的代码都超时了，就不贴出来了-_-||| 后来参照leetcode的讨论，发现有人提出用两个HashMap分别存储A、B和C、D任意两个元素的和，若第一个map中的某个key，它在相反数（-key）在第二个map中存在，则组成它们的4个数组中的元素的和就是0，而对于本组的4元组的个数，则为key在第一个map中的次数（即key对应的值）与-key在第二个map中的次数。最终代码如下：

    int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {
        unordered_map<int, int> mp, mp1;
        for (int a : A)
            for (int b : B)
                mp[a + b]++;
        for (int c : C)
            for (int d : D)
                mp1[c + d]++;
        int rst = 0, tmp = 0;
        for (auto iter : mp)
        {
            tmp = -1 * iter.first;
            auto it = mp1.find(tmp);
            if (it != mp1.end())
                rst += iter.second * it->second;
        }
        return rst;
    }

标签：4Sum,map,int,454,II,vector,key,mp1,rst
来源： https://www.cnblogs.com/big-potato/p/12269233.html