[Locked] 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:

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 |

分析：

　　第1种，求解一般矩阵乘积的方法；第2种，根据稀疏矩阵的特性减少0*x的计算次数。

代码1：

public:
vector<vector<int>> multiply(vector<vector<int>>& A, vector<vector<int>>& B) {
int m = A.size(), n = B.size(), p = B[0].size();
vector<vector<int> >C(m, vector<int>(p, 0));
for(int i = 0; i < m; i++) {
for(int j = 0; j < n; j++) {
if(A[i][j]) {
for(int k = 0; k < p; k++) {
C[i][k] += A[i][j] * B[j][k];
}
}
}
}
return C;
}
};

代码2：

待补充