LintCode 654 · Sparse Matrix Multiplication (稀疏矩阵乘法)

654 · Sparse Matrix Multiplication
Algorithms

Description
Given two Sparse Matrix A and B, return the result of AB.

You may assume that A’s column number is equal to B’s row number.

Example
Example1

Input:
[[1,0,0],[-1,0,3]]
[[7,0,0],[0,0,0],[0,0,1]]
Output:
[[7,0,0],[-7,0,3]]
Explanation:
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 |
Example2

Input:
[[1,0],[0,1]]
[[0,1],[1,0]]
Output:
[[0,1],[1,0]]

解法1：常规解法。没用到稀疏矩阵的优势。

class Solution {
public:
    /**
     * @param a: a sparse matrix
     * @param b: a sparse matrix
     * @return: the result of A * B
     */
    vector<vector<int>> multiply(vector<vector<int>> &a, vector<vector<int>> &b) {
        int m = a.size();
        int n = b.size();
        if (m == 0 || n == 0) return vector<vector<int>>();
        int p = b[0].size();
        vector<vector<int>> res(m, vector<int>(p, 0));
        for (int i = 0; i < m; i++) {
            for (int k = 0; k < p; k++) {
                int sum = 0;
                for (int j = 0; j < n; j++) {
                    sum += a[i][j] * b[j][k]; 
                }
                res[i][k] = sum;
            }
            
        }
        return res;
    }
};

解法2: 利用了稀疏矩阵的部分优势，也就是当a[i][j] == 0时，省掉k循环。

class Solution {
public:
    /**
     * @param a: a sparse matrix
     * @param b: a sparse matrix
     * @return: the result of A * B
     */
    vector<vector<int>> multiply(vector<vector<int>> &a, vector<vector<int>> &b) {
        int m = a.size();
        int n = b.size();
        if (m == 0 || n == 0) return vector<vector<int>>();
        int p = b[0].size();
        vector<vector<int>> res(m, vector<int>(p, 0));
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (a[i][j] == 0) continue;
                for (int k = 0; k < p; k++) {
                    res[i][k] += a[i][j] * b[j][k]; 
                }
            }
        }
        return res;
    }
};