Ashes To Glory: Rotating a 2D array of integers (matrix) by a given angle (+90, -90, +180, -180)

Related: Inplace M x N size matrix transpose
You are given a 2D square matrix, or 2D array of integers of size n (n rows and n columns), your output should be n by n 2D matrix rotated by a given angle, which could be +90, -90, +180, -180.

Rotate by +90 (clockwise once):

Input: n by n matrix M, where n >= 2

Algorithm:

Step 1: Transpose M

Step 2: Reverse each row

Dry run:

Step 1: Transpose

M                   M'

1 2 3              1 4 7

4 5 6      --     2 5 8

7 8 9              3 6 9

Step 2: Reverse each row

M'                  M''

1 4 7              7 4 1
2 5 8      --     8 5 2
3 6 9              9 6 3

Pseudocode: is here which is self explanatory and easily convertible to source code in a language of your choice.

Transpose

    for i in [0, n)

        for j in [0, n)

            if ( i < j )

                swap( M[i][j], M[j][i] )

Reverse a row (rowidx)

    start = 0

    end = cols - 1

    while ( start < end ) {

        swap( M[rowidx][start], M[rowidx][end] )

        ++start

        --end

    }

Rotate

    Transpose

    for i in [0, rows)

        Reverse( i )

That was fair enough until only asked to rotate by +90 degree. If problem is further extended to be solved for any given angle, of course the rotation should make sense, for example, 47 degree is not a choice ;-). So, here are some elegant techniques (I'll be brief now for other angles, because for +90 degree I have elaborated the solution):

Rotation by -90 degree (anticlockwise once):

Step 1: Transpose

Step 2: Reverse each column

Rotation by +180 degree (clockwise twice): Two methods follows

First:

Rotate input matrix +90 degree twice, if routine for which is available to you

Second: (You'll be amazed!)

Step 1: Reverse each row

Step 2: Reverse each column

Rotation by -180 degree (anticlockwise twice): Three(!!!) methods follows

First:

Rotate input matrix -90 degree twice, if routine for which is available to you

Second: (You'll be amazed again!)

Step 1: Reverse each column

Step 2: Reverse each row

Third: (Aha!)

Because rotating a matrix +180 degree or -180 should produce same result. So you can rotate it by +180 degree using one of above methods.

Turn an image by 90 degree

From http://www.geeksforgeeks.org/turn-an-image-by-90-degree/

for(r = 0; r < m; r++)
{
   for(c = 0; c < n; c++)
   {
      // Hint: Map each source element indices into
      // indices of destination matrix element.
       dest_buffer [ c ] [ m - r - 1 ] = source_buffer [ r ] [ c ];
   }
}

Code From http://www.emanueleferonato.com/2012/11/07/how-to-rotate-a-two-dimensional-array-by-90-degrees-clockwise-or-counter-clockwise-like-knightfall-game/

        private function rotateCounterClockwise(a:Array):void {

            var n:int=a.length;

            for (var i:int=0; i<n/2; i++) {

                for (var j:int=i; j<n-i-1; j++) {

                    var tmp:String=a[i][j];

                    a[i][j]=a[j][n-i-1];

                    a[j][n-i-1]=a[n-i-1][n-j-1];

                    a[n-i-1][n-j-1]=a[n-j-1][i];

                    a[n-j-1][i]=tmp;

                }

            }

        }

        private function rotateClockwise(a:Array):void {

            var n:int=a.length;

            for (var i:int=0; i<n/2; i++) {

                for (var j:int=i; j<n-i-1; j++) {

                    var tmp:String=a[i][j];

                    a[i][j]=a[n-j-1][i];

                    a[n-j-1][i]=a[n-i-1][n-j-1];

                    a[n-i-1][n-j-1]=a[j][n-i-1];

                    a[j][n-i-1]=tmp;

                }

            }

        }

https://rosettacode.org/wiki/Matrix_transposition#Java

               double[][] ans = new double[m[0].length][m.length];
               for(int rows = 0; rows < m.length; rows++){
                       for(int cols = 0; cols < m[0].length; cols++){
                               ans[cols][rows] = m[rows][cols];
                       }
               }

http://introcs.cs.princeton.edu/java/14array/Transpose.java.html

        // transpose in-place
        for (int i = 0; i < N; i++) {
            for (int j = i+1; j < N; j++) {
                int temp = a[i][j];
                a[i][j] = a[j][i];
                a[j][i] = temp;
            }
        }

http://analgorithmaday.blogspot.com/2011/04/rotate-array90.html
http://coding-interview-archives.blogspot.com/2013/09/rotate-given-square-matrix-by-90-degrees.html
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