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[].size();
vector<vector<int> >C(m, vector<int>(p, ));
for(int i = ; i < m; i++) {
for(int j = ; j < n; j++) {
if(A[i][j]) {
for(int k = ; k < p; k++) {
C[i][k] += A[i][j] * B[j][k];
}
}
}
}
return C;
}
};
代码2：
待补充