[leetcode]311. Sparse Matrix Multiplication 稀疏矩阵相乘

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 |

注意：

搞清楚何谓matrix multiply:

一定要有A column 等于B row的特性才能进行matrix multiply

     |  1 0 0 |   | 7 0 0 |   |  7 0 0 |   //  1*7 + 0*0 + 0*0 = 7
AB = | -1 0 3 | x | 0 0 0 | = | -7 0 3 |
                  | 0 0 1 |

     |  1 0 0 |   | 7 0 0 |   |  7 0 0 |   //  1*0 + 0*0 + 0*0 = 0
AB = | -1 0 3 | x | 0 0 0 | = | -7 0 3 |
                  | 0 0 1 |

     |  1 0 0 |   | 7 0 0 |   |  7 0 0 |  // 1*0 + 0*0 + 0*1 = 0
AB = | -1 0 3 | x | 0 0 0 | = | -7 0 3 |
                  | 0 0 1 |

     |  1 0 0 |   | 7 0 0 |   |  7 0 0 |
AB = | -1 0 3 | x | 0 0 0 | = | -7 0 3 | // -1*7 + 0*0 + 3*0 = -7
                  | 0 0 1 |

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思路：

Brute Force:  create product 2D matrix, iterate through it and calculate result for each position

Optimized: Use the information that matrix is sparse. Iterate through A and add the contribution of each number to the result matrix. If A[i][j] == 0, skip the calculation

代码：

 1 class Solution {
 2     public int[][] multiply(int[][] A, int[][] B) {
 3         int m =  A.length, n = A[0].length;
 4         int nB = B[0].length;
 5         int [][] res = new int[m][nB];
 6         
 7         for(int i = 0; i< m; i++){
 8             for(int k = 0; k < n; k++){
 9                 if(A[i][k]!=0){ // use Sparse Matrix attributes
10                     for(int j = 0; j < nB; j++){
11                         if(B[k][j]!=0) res[i][j] += A[i][k] *B[k][j];                    
12                     }
13                 }
14             }
15         }
16        return res;  
17     }  
18 }