Rotating a 2D pixel array by 90 degrees

This can be done without using any extra space, so called In-place matrix transposition (not exact the same). Remember to do some mirroring after the transposition.

1. If the image is square

   

2. If the image is not square

   • For non-square matrices, the algorithms are more complicated. Many of the algorithms prior to 1980 could be described as "follow-the-cycles" algorithms. That is, they loop over the cycles, moving the data from one location to the next in the cycle. In pseudocode form:

   

You have old_data[rows][cols] and new_data[cols][rows], then:

for(int i=0; i<cols; i++) {
    for(int j=0; j<rows; j++) {
        new_data[i][j] = old_data[rows-1-j][i];
    }
}

This should rotate old_data by 90 degrees CW.

If you want to do it in-place with O(1) space, you can follow this:

1. Transpose the matrix by swapping data[i][j] and data[j][i] :

   for (int i = 0; i < n; i += 1) {
       for (int j = i+1; j < n; j += 1) {
           swap(data[i][j], data[j][i]);
       }
   }
   

2. Reverse each row or column for +90 or -90 degrees of rotation, respectively. For example for +90 degrees of rotation:

   for (int i = 0; i < n; i += 1) {
       for (int j = 0; j < n/2; j += 1) {
           swap(data[i][j], data[i][n-1-j]);
       }
   }