Given two sparse matrices A and B, return the result of AB.
You may assume that A‘s column number is equal to B‘s row number.
Example:
A = [ [ 1, 0, 0], [-1, 0, 3] ] B = [ [ 7, 0, 0 ], [ 0, 0, 0 ], [ 0, 0, 1 ] ] | 1 0 0 | | 7 0 0 | | 7 0 0 | AB = | -1 0 3 | x | 0 0 0 | = | -7 0 3 | | 0 0 1 |
分析:
第1种,求解一般矩阵乘积的方法;第2种,根据稀疏矩阵的特性减少0*x的计算次数。
代码1:
public: vector<vector<int>> multiply(vector<vector<int>>& A, vector<vector<int>>& B) { int m = A.size(), n = B.size(), p = B[0].size(); vector<vector<int> >C(m, vector<int>(p, 0)); for(int i = 0; i < m; i++) { for(int j = 0; j < n; j++) { if(A[i][j]) { for(int k = 0; k < p; k++) { C[i][k] += A[i][j] * B[j][k]; } } } } return C; } };
代码2:
待补充
时间: 2024-07-30 03:23:07