Leetcode.454.4SumII

作者: Jimmy木 | 来源:发表于2020-01-07 20:40 被阅读0次

题目

给定4个等长的整型数组, 每个数组取一个数组结果为0, 输出一共有多少种可能.例如A, B, C, D, A[a] + B[b] + C[c] + D[d] = 0.

Input: A = [ 1, 2]
       B = [-2,-1]
       C = [-1, 2]
       D = [ 0, 2]
Output: 2
Explanation: 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

思路

先用map记录两个数组的和, 然后计算剩下两个数组和, 以之前的map比较.

int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {
    int res = 0;
    unordered_map<int, int> ab;
    for (int a : A) {
        for (int b : B) {
            ab[a+b]++;
        }
    }

    for (int c : C) {
        for (int d : D) {
            if (ab.count(-c-d) > 0) {
                res += ab[-c-d];
            }
        }
    }
    return res;
}