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:

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 |

方法1:

思路：

主要是需要sparse matrix里大量的0 产生的无用计算。那么什么样的计算才会真正影响结果呢？当A[i][k] !=0 && B[k][j] != 0。那么在每次循环之前检查一下，剪枝掉所有不需要的计算：任何一个数等于0就不需要进入循环。

Complexity

Time complexity: O(mkn)
Space comlexity: O(mn)

class Solution {
public:
    vector<vector<int>> multiply(vector<vector<int>>& A, vector<vector<int>>& B) {
        vector<vector<int>> result(A.size(), vector<int>(B[0].size(), 0));
        for (int i = 0 ; i < A.size(); i++){
            for (int k = 0; k < A[0].size(); k++){
                if (A[i][k] != 0) {
                    for (int j = 0 ; j < B[0].size(); j++){
                        if (B[k][j] != 0) result[i][j] += A[i][k] * B[k][j];
                    }
                }
            }
        }
        return result;
    }
};