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:
Input:
A = [
[ 1, 0, 0],
[-1, 0, 3]
]
B = [
[ 7, 0, 0 ],
[ 0, 0, 0 ],
[ 0, 0, 1 ]
]
Output:
| 1 0 0 | | 7 0 0 | | 7 0 0 |
AB = | -1 0 3 | x | 0 0 0 | = | -7 0 3 |
| 0 0 1 |
代码
Approach #1
classSolution {publicint[][] multiply(int[][] A,int[][] B) {int m =A.length, n =A[0].length;int nB =B[0].length;int[][] C =newint[m][nB];for (int i =0; i < m; i++) {for (int k =0; k < n; k++) {if (A[i][k] !=0) {for (int j =0; j < nB; j++) {if (B[k][j] !=0) C[i][j] +=A[i][k] *B[k][j]; } } } }return C; }}
classSolution {publicint[][] multiply(int[][] A,int[][] B) {int m =A.length, n =A[0].length, nB =B[0].length;int[][] C =newint[m][nB];for (int i =0; i < m; i++) {for (int j =0; j < nB; j++) {for (int k =0; k < n; k++) {C[i][j] +=A[i][k] *B[k][j]; } } }return C; }}