Sparse Matrix Multiplication
題意：
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 |
解題思路：
暴力法，會超時：
public class Solution {
public int[][] multiply(int[][] A, int[][] B) {
int rowA = A.length;
int colA = A[0].length;
int rowB = B.length;
int colB = B[0].length;
int[][] res = new int[rowA][colB];
for (int i = 0; i < rowA; i++) {
for (int j = 0; j < colB; j++) {
for (int k = 0; k < colB; k++) {
res[i][j] += A[i][k] * B[k][j];
}
}
}
return res;
}
}
因是sparse matrix，必包含非常多的0 ，因此我們用column major的一一判斷，一但A[i][k]為0 的話，後面也不需要作了，可以省去非常多的時間，其程式碼如下：
public class Solution {
public int[][] multiply(int[][] A, int[][] B) {
int rowA = A.length;
int colA = A[0].length;
int rowB = B.length;
int colB = B[0].length;
int[][] res = new int[rowA][colB];
for (int i = 0; i < rowA; i++) {
for (int k = 0; k < colA; k++) {
if (A[i][k] != 0) {
for (int j = 0; j < colB; j++) {
res[i][j] += A[i][k] * B[k][j];
}
}
}
}
return res;
}
}