# Sparse Matrix Multiplication

### 描述​

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 |

Constraints:

• 1 <= A.length, B.length <= 100
• 1 <= A[i].length, B[i].length <= 100
• -100 <= A[i][j], B[i][j] <= 100

### 代码​

// Sparse Matrix Multiplication// Time Complexity: O(m*n*p), Space Complexity: O(1)public class Solution {    public int[][] multiply(int[][] A, int[][] B) {        int m = A.length, n = A[0].length, p = B[0].length;        int[][] C = new int[m][p];        for(int i = 0; i < m; i++) {            for(int k = 0; k < n; k++) {                if (A[i][k] != 0) {                    for (int j = 0; j < p; j++) {                        if (B[k][j] != 0) C[i][j] += A[i][k] * B[k][j];                    }                }            }        }        return C;    }}