为新手程序员，在纯c++。(宝蓝的东西)

#include<iostream.h>
#include<conio.h>

int main()
{
    clrscr();

    int arr[10][10];        // 2d array that holds input elements 
    int result[10][10];     //holds result

    int m,n;                //rows and columns of arr[][]
    int x,y;                //rows and columns of result[][]

    int i,j;                //loop variables
    int t;                  //temporary , holds data while conversion

    cout<<"Enter no. of rows and columns of array: ";
    cin>>m>>n;
    cout<<"\nEnter elements of array: \n\n";
    for(i = 0; i < m; i++)
    {
        for(j = 0; j<n ; j++)
        {
          cin>>arr[i][j];         // input array elements from user
        }
    }


   //rotating matrix by +90 degrees

    x = n ;                      //for non-square matrix
    y = m ;     

    for(i = 0; i < x; i++)
    {  t = m-1;                     // to create required array bounds
       for(j = 0; j < y; j++)
       {
          result[i][j] = arr[t][i];
          t--;
       }
   }

   //print result

   cout<<"\nRotated matrix is: \n\n";
   for(i = 0; i < x; i++)
   {
       for(j = 0; j < y; j++)
       {
             cout<<result[i][j]<<" ";
       }
       cout<<"\n";
   }

   getch();
   return 0;
}

在python中:

import numpy as np

a = np.array(
    [
        [1, 2, 3, 4],
        [5, 6, 7, 8],
        [9, 0, 1, 2],
        [3, 4, 5, 6]
    ]
)

print(a)
print(b[::-1, :].T)

private static int[][] rotate(int[][] matrix, int n) {
    int[][] rotated = new int[n][n];
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            rotated[i][j] = matrix[n-j-1][i];
        }
    }
    return rotated;
}

下面是PHP的递归方法:

$m = array();
            $m[0] = array('a', 'b', 'c');
            $m[1] = array('d', 'e', 'f');
            $m[2] = array('g', 'h', 'i');
            $newMatrix = array();

            function rotateMatrix($m, $i = 0, &$newMatrix)
            {
                foreach ($m as $chunk) {
                    $newChunk[] = $chunk[$i];
                }
                $newMatrix[] = array_reverse($newChunk);
                $i++;

                if ($i < count($m)) {
                    rotateMatrix($m, $i, $newMatrix);
                }
            }

            rotateMatrix($m, 0, $newMatrix);
            echo '<pre>';
            var_dump($newMatrix);
            echo '<pre>';

Python:

rotated = list(zip(*original[::-1]))

和逆时针方向:

rotated_ccw = list(zip(*original))[::-1]

这是如何工作的:

Zip (*original)将通过将列表中的对应项堆叠到新的列表中来交换2d数组的轴。(*操作符告诉函数将包含的列表分布到参数中)

>>> list(zip(*[[1,2,3],[4,5,6],[7,8,9]]))
[[1,4,7],[2,5,8],[3,6,9]]

语句[::-1]反转数组元素(请参阅扩展切片或这个问题):

>>> [[1,2,3],[4,5,6],[7,8,9]][::-1]
[[7,8,9],[4,5,6],[1,2,3]]

最后，将两者结合就得到了旋转变换。

改变[::-1]的位置将使列表在矩阵的不同层次上颠倒。

Nick的答案也适用于NxM阵列，只需要做一点修改(与NxN相反)。

string[,] orig = new string[n, m];
string[,] rot = new string[m, n];

...

for ( int i=0; i < n; i++ )
  for ( int j=0; j < m; j++ )
    rot[j, n - i - 1] = orig[i, j];

考虑这个问题的一种方法是将轴(0,0)的中心从左上角移动到右上角。你只是简单地从一个转置到另一个。