311. Sparse Matrix Multiplication

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 |

本题是稀疏矩阵，稀疏矩阵的特点是有很多0，因此不能用传统的计算方法来进行运算，这样会TLE，假设通式是这样的C[i][j] += A[i][k]*B[k][j];（不会解释。。。。。）代码如下：

 1 public class Solution {
 2     public int[][] multiply(int[][] A, int[][] B) {
 3         int row = A.length;
 4         int col = B[0].length;
 5         int z = A[0].length;
 6         int[][] C = new int[row][col];
 7         for(int i=0;i<row;i++){
 8             for(int k=0;k<z;k++){
 9                 if(A[i][k]!=0){
10                     for(int j=0;j<col;j++){
11                         if(B[k][j]!=0) C[i][j]+=A[i][k]*B[k][j]; 
12                     }
13                 }
14             }
15         }
16         return C;
17     }
18 }