311. Sparse-Matrix-Multiplication

difficulty: Medium

section pre{ background-color: #eee; border: 1px solid #ddd; padding:10px; border-radius: 5px; }

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 |

Constraints:

• 1 <= A.length, B.length <= 100

• 1 <= A[i].length, B[i].length <= 100

• -100 <= A[i][j], B[i][j] <= 100

Method One

class Solution {
    public int[][] multiply(int[][] A, int[][] B) {
        // 不停剪枝就好了。其实就是普通乘法多了两个判断, 有0那我整个就不算了。
        int mA = A.length;
        int nA = A[0].length;
        int mA = B.length; // if nA != mA 要报错。
        int nB = B[0].length;

        int[][] ans = new int[mA][nB];

        for(int rowA = 0; rowA < mA; rowA++ ) {
            // index 其实就是 colA, 也等于 rowB.
            for(int index = 0; index < nA; index++ ) {
                if( A[rowA][index] == 0) {
                    continue;
                }
                for(int colB = 0; colB < nB; colB++ ) {
                    if(B[index][colB] == 0) {
                        continue;
                    }
                    int val = A[rowA][index] * B[index][colB];
                    ans[rowA][colB] += val;
                }
            }
        }
        return ans;
    }
}
​