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

分析

无

代码

Python

// 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;
    }
}

Java

// TODO

C++

# Sparse Matrix Multiplication
# Time Complexity: O(m*n*p), Space Complexity: O(1)
def multiply(A: list[list[int]], B: list[list[int]]) -> list[list[int]]:
    m, n, p = len(A), len(A[0]), len(B[0])
    C = [[0] * p for _ in range(m)]

    for i in range(m):
        for k in range(n):
            if A[i][k] != 0:
                for j in range(p):
                    if B[k][j] != 0:
                        C[i][j] += A[i][k] * B[k][j]
    return C