力扣（LeetCode）算法_C++——稀疏矩阵的乘法

给定两个 稀疏矩阵 ：大小为 m x k 的稀疏矩阵 mat1 和大小为 k x n 的稀疏矩阵 mat2 ，返回 mat1 x mat2 的结果。你可以假设乘法总是可能的。

示例 1：

输入：mat1 = \[1,0,0,-1,0,3], mat2 = \[7,0,0,0,0,0,0,0,1]

输出：\[7,0,0,-7,0,3]

示例 2:

输入：mat1 = \[0], mat2 = \[0]

输出：\[0]

    vector<vector<int>> multiply(vector<vector<int>>& mat1, vector<vector<int>>& mat2) {
        int m = mat1.size();
        int k = mat1[0].size();
        int n = mat2[0].size();
        unordered_map<int, list<pair<int, int>>> map1;
        unordered_map<int, list<pair<int, int>>> map2;
        
        for (int i {0}; i < m; i++) {
            for (int j {0}; j < k; j++) {
                if (mat1[i][j] != 0) map1[mat1[i][j]].push_back(make_pair(i, j));
            }
        }        
        
        for (int i = 0; i < k; i++) {
            for (int j = 0; j < n; j++) {
                if (mat2[i][j] != 0) map2[mat2[i][j]].push_back(make_pair(i, j));
            }
        }

        vector<vector<int>> res(m, vector<int>(n, 0));
        if (m == 0 || n == 0) return res;

        for (auto& element1 : map1) {
            for (auto& list_node1 : element1.second) {
                for (auto& element2 : map2) {
                    for (auto& list_node2 : element2.second) {
                        if (list_node1.second == list_node2.first) {
                            res[list_node1.first][list_node2.second] += element1.first * element2.first;
                        }
                    }
                }
            }
            
        }

        return res;
    }