Sparse Matrix Multiplication

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:

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 |

Solution

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

//  以其中一个数组A为基准循环，这样可以先检查A［i］［j］不为零才向下继续循环。

Solution 2 

用hash map 将第三个循环优化下，只记录 第k 行， j 列不为零的树值。

A sparse matrix can be represented as a sequence of rows, each of which is a sequence of (column-number, value) pairs of the nonzero values in the row.

public class Solution {
    public int[][] multiply(int[][] A, int[][] B) {
        if (A == null || A[0] == null || B == null || B[0] == null) return null;
        int m = A.length, n = A[0].length, l = B[0].length;
        int[][] C = new int[m][l];
        Map<Integer, HashMap<Integer, Integer>> tableB = new HashMap<>();

        for(int k = 0; k < n; k++) {
            tableB.put(k, new HashMap<Integer, Integer>());
            for(int j = 0; j < l; j++) {
                if (B[k][j] != 0){
                    tableB.get(k).put(j, B[k][j]);
                }
            }
        }

        for(int i = 0; i < m; i++) {
            for(int k = 0; k < n; k++) {
                if (A[i][k] != 0){
                    for (Integer j: tableB.get(k).keySet()) {
                        C[i][j] += A[i][k] * tableB.get(k).get(j);
                    }
                }
            }
        }
        return C;   
    }
}

Solution 2 refer : http://www.cnblogs.com/EdwardLiu/p/5090563.html