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 | 根据矩阵乘法的规则写程序，不过要以A、B矩阵元素为遍历单位，这样可以跳过为0的元素。代码如下： public class Solution { public int[][] multiply(int[][] A, int[][] B) { if (A.length == 0 || A[0].length == 0 || B.length == 0 || B[0].length == 0) { return null; } int[][] res = new int[A.length][B[0].length]; for (int i = 0; i < A.length; i ++) { for (int j = 0; j < A[0].length; j ++) { if (A[i][j] != 0) { for (int k = 0; k < B[0].length; k ++) { if (B[j][k] != 0) { res[i][k] += A[i][j] * B[j][k]; } } } } } return res; } }