```
class Solution {
public:
void rotate(vector<vector<int> > &matrix) {
int n = matrix.size();
int lo = 0, hi = n-2;
while(lo <= hi){
doRotation(matrix, lo, hi, n);
lo++;
hi--;
}
return;
}
private:
void doRotation(vector<vector<int> > &matrix, int p, int q, int n){
int tmp;
for(int j = p; j<=q; j++){
tmp = matrix[p][j];
matrix[p][j] = matrix[n-1-j][p];
matrix[n-1-j][p] = matrix[n-1-p][n-1-j];
matrix[n-1-p][n-1-j] = matrix[j][n-1-p];
matrix[j][n-1-p] = tmp;
}
return;
}
};
```

I treat this matrix as a swirl, and start from the outskirt.

For the most outer level, we only need to deal with the 0 to n-2 entries of each edge.

Then we do the same thing from level 0 to level n/2-1. It's simple, linear time and in place :)