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## Problem

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 |


## Code

class Solution {
public int[][] multiply(int[][] A, int[][] B) {
int rowA = A.length;
int colA = A[0].length; // colA == rowB
int colB = B[0].length;

int[][] res = new int[rowA][colB];

for(int i = 0; i < rowA; i++){
for(int j = 0; j < colA; j++){
if(A[i][j] != 0){
for(int k = 0; k < colB; k++){
if(B[j][k] != 0){
// res[i][k] = Aij * Bjk + Aij * Bjk + Aij * Bjk
// res[0][0] = A00 * B00 + A01 * B10 + A02 * B20
// i和k始终是0， j不停的变换0，1，2
res[i][k] += A[i][j] * B[j][k];
}
}
}
}
}
return res;
}
}

class Solution:
def multiply(self, A: List[List[int]], B: List[List[int]]) -> List[List[int]]:
if len(A) == 0 or len(B) == 0:
return [[]]

a, b, c = len(A), len(B), len(B[0])
res = [[0 for _ in range(c)] for _ in range(a)]

for i in range(a):
for j in range(b):
if A[i][j] != 0:
for k in range(c):
if B[j][k] != 0:
res[i][k] += A[i][j] * B[j][k]

return res