[LeetCode]Rotate the image(在位操作！！）

You are given an n x n 2D matrix representing an image.

Rotate the image by 90 degrees (clockwise).

Follow up: Could you do this in-place?

在位操作说明不能开辟新的空间

void rotate(int** matrix, int r, int c) {
int **p = NULL;
int i,j;
p = (int**)malloc(sizeof(int*)*c);
for(i = 0; i < c; ++i)
p[i] =(int*)malloc(sizeof(int)*r);

for(i = 0; i < c; ++i)
for(j = 0; j<r;++j)
p[i][j]=matrix[i][j];
for(i = 0; i < c; ++i)
for(j = 0; j<r;++j)
//p[i][j]=matrix[r-1-j][i];
matrix[i][j]=p[r-1-j][i];

}

这种方法虽然0msAC，但是明显的开辟了空间

这里有个新的方法

/*
* clockwise rotate
* first reverse up to down, then swap the symmetry
* 1 2 3     7 8 9     7 4 1
* 4 5 6  => 4 5 6  => 8 5 2
* 7 8 9     1 2 3     9 6 3
*/
void rotate(vector<vector<int> > &matrix) {
reverse(matrix.begin(), matrix.end());
for (int i = 0; i < matrix.size(); ++i) {
for (int j = i + 1; j < matrix[i].size(); ++j)
swap(matrix[i][j], matrix[j][i]);
}
}

/*
* anticlockwise rotate
* first reverse left to right, then swap the symmetry
* 1 2 3     3 2 1     3 6 9
* 4 5 6  => 6 5 4  => 2 5 8
* 7 8 9     9 8 7     1 4 7
*/
void anti_rotate(vector<vector<int> > &matrix) {
for (auto vi : matrix) reverse(vi.begin(), vi.end());
for (int i = 0; i < matrix.size(); ++i) {
for (int j = i + 1; j < matrix[i].size(); ++j)
swap(matrix[i][j], matrix[j][i]);
}
}