Question

Formatted question description: https://leetcode.ca/all/311.html

 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 |

Algorithm

Make sure that A[i][k] is not 0 before continuing to calculate, and then traverse the kth row of matrix B. If B[K][J] is not 0, accumulate the result matrix res[i][j] += A[i][k] * B[k][j], so that we can efficiently calculate the multiplication of the sparse matrix.

Code

Java

public class Sparse_Matrix_Multiplication {

    public int[][] multiply(int[][] A, int[][] B) {

        //validity check
        int[][] C = new int[A.length][B[0].length];

        for (int i = 0; i < C.length; i++) {
            for (int k = 0; k < A[0].length; k++) {
                if (A[i][k] != 0) { // @note: non-zero check. if zero, then skip since result will be 0
                    for (int j = 0; j < C[0].length; j++) {
                        C[i][j] += A[i][k] * B[k][j];
                    }
                }
            }
        }

        return C;
    }
}

// OJ: https://leetcode.com/problems/sparse-matrix-multiplication/
// Time: O(MNK)
// Space: O(1)
class Solution {
public:
    vector<vector<int>> multiply(vector<vector<int>>& A, vector<vector<int>>& B) {
        int M = A.size(), K = A[0].size(), N = B[0].size();
        vector<vector<int>> ans(M, vector<int>(N));
        for (int i = 0; i < M; ++i) {
            for (int j = 0; j < N; ++j) {
                for (int k = 0; k < K; ++k) {
                    ans[i][j] += A[i][k] * B[k][j];
                }
            }
        }
        return ans;
    }
};

class Solution(object):
  def multiply(self, A, B):
    """
    :type A: List[List[int]]
    :type B: List[List[int]]
    :rtype: List[List[int]]
    """
    ret = [[0 for j in range(len(B[0]))] for i in range(len(A))]

    for i, row in enumerate(A):
      for k, a in enumerate(row):
        if a:
          for j, b in enumerate(B[k]):
            if b:
              ret[i][j] += a * b
    return ret

311 - Sparse Matrix Multiplication

Question

Algorithm

Code

All Problems

All Solutions