Leetcode: 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 |

70ms solution:

public class Solution {
    public int[][] multiply(int[][] A, int[][] B) {
        int m = A.length; 
        int n = A[0].length;
        int bn = B[0].length;
        int[][] C = new int[m][bn];
        
        for (int i=0; i<m; i++) {
            for (int k=0; k<n; k++) {
                if (A[i][k] == 0) continue;
                for (int j=0; j<bn; j++) {
                    C[i][j] += A[i][k] * B[k][j];
                }
            }
        }
        return C;
    }
}

use one HashTable: 160 ms

The idea is derived from a CMU lecture.

A sparse matrix can be represented as a sequence of rows, each of which is a sequence of (column-number, value) pairs of the nonzero values in the row.

public class Solution {
    public int[][] multiply(int[][] A, int[][] B) {
        if (A == null || A[0] == null || B == null || B[0] == null) return null;
        int m = A.length, n = A[0].length, l = B[0].length;
        int[][] C = new int[m][l];
        Map<Integer, HashMap<Integer, Integer>> tableB = new HashMap<>();

        for(int k = 0; k < n; k++) {
            tableB.put(k, new HashMap<Integer, Integer>());
            for(int j = 0; j < l; j++) {
                if (B[k][j] != 0){
                    tableB.get(k).put(j, B[k][j]);
                }
            }
        }

        for(int i = 0; i < m; i++) {
            for(int k = 0; k < n; k++) {
                if (A[i][k] != 0){
                    for (Integer j: tableB.get(k).keySet()) {
                        C[i][j] += A[i][k] * tableB.get(k).get(j);
                    }
                }
            }
        }
        return C;   
    }
}