A straightforward solution using C++, using O(m+n) space.

The basic idea is record which row and column needs to be set zero then change the numbers one by one.

The time complexity is O(m*n)

```
public:
void setZeroes(vector<vector<int>>& matrix) {
int m = matrix.size();
if(m == 0) return;
int n = matrix[0].size();
vector<bool> row(m, true);
vector<bool> col(n, true);
for(int i = 0; i < m; i++) {
for(int j = 0; j < n; j++) {
if(matrix[i][j] == 0) {
row[i] = false;
col[j] = false;
}
}
}
for(int i = 0; i < m; i++) {
for(int j = 0; j < n; j++) {
if(!row[i] || !col[j]) {
matrix[i][j] = 0;
}
}
}
return;
}
};
```