leetcode311- Sparse Matrix Multiplication- medium

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 |

 1.O(mkn)复杂度。三重for循环直接做。

2.O(mk + nk)复杂度。Map<Integer, List<Integer>>或者List<List<Integer>>来辅助做。存储B矩阵里面非0数的横纵坐标B[k][j]信息。之后遍历A矩阵，入股遇到非0数A[i][k]，用存储的信息做一次for循环。非0数A[i][k]就是最后能做贡献的数，它最后会和所有共享k的B[k][...]的数一起做贡献，具体和B[k][j]数就是贡献到result[i][j]的位置。

 1.O(mkn)复杂度

class Solution {
    public int[][] multiply(int[][] A, int[][] B) {
        if (A == null || B == null || A.length == 0 || A[0].length == 0 || B[0].length == 0) {
            return new int[0][0];
        }
        int m = A.length;
        int c = A[0].length;
        int n = B[0].length;
        int[][] result = new int[m][n];
        
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                int sum = 0;
                for (int k = 0; k < c; k++) {
                    sum += A[i][k] * B[k][j];
                }
                result[i][j] = sum;
            }
        }
        
        return result;
    }
}

2.O(mk+nk)复杂度。

class Solution {
    public int[][] multiply(int[][] A, int[][] B) {
        if (A == null || B == null || A.length == 0 || A[0].length == 0 || B[0].length == 0) {
            return new int[0][0];
        }
        int m = A.length;
        int t = A[0].length;
        int n = B[0].length;
        int[][] result = new int[m][n];
        Map<Integer, List<Integer>> idxB = new HashMap<>();
        
        for (int k = 0; k < t; k++) {
            idxB.put(k, new ArrayList<Integer>());
            for (int j = 0; j < n; j++) {
                if (B[k][j] != 0) {
                    idxB.get(k).add(j);
                }
            }
        }
        
        for (int i = 0; i < m; i++) {
            for (int k = 0; k < t; k++) {
                if (A[i][k] == 0) {
                    continue;
                }
                for (int j : idxB.get(k)) {
                    result[i][j] += A[i][k] * B[k][j];
                }
            }
        }
        
        return result;
    }
}