你可以通过3个简单步骤做到这一点:

1)假设我们有一个矩阵

   1 2 3
   4 5 6
   7 8 9

2)求矩阵的转置

   1 4 7
   2 5 8
   3 6 9

3)交换行得到旋转矩阵

   3 6 9
   2 5 8
   1 4 7

Java源代码:

public class MyClass {

    public static void main(String args[]) {
        Demo obj = new Demo();
        /*initial matrix to rotate*/
        int[][] matrix = { { 1, 2, 3 }, { 4, 5, 6 }, { 7, 8, 9 } };
        int[][] transpose = new int[3][3]; // matrix to store transpose

        obj.display(matrix);              // initial matrix

        obj.rotate(matrix, transpose);    // call rotate method
        System.out.println();
        obj.display(transpose);           // display the rotated matix
    }
}

class Demo {   
    public void rotate(int[][] mat, int[][] tran) {

        /* First take the transpose of the matrix */
        for (int i = 0; i < mat.length; i++) {
            for (int j = 0; j < mat.length; j++) {
                tran[i][j] = mat[j][i]; 
            }
        }

        /*
         * Interchange the rows of the transpose matrix to get rotated
         * matrix
         */
        for (int i = 0, j = tran.length - 1; i != j; i++, j--) {
            for (int k = 0; k < tran.length; k++) {
                swap(i, k, j, k, tran);
            }
        }
    }

    public void swap(int a, int b, int c, int d, int[][] arr) {
        int temp = arr[a][b];
        arr[a][b] = arr[c][d];
        arr[c][d] = temp;    
    }

    /* Method to display the matrix */
    public void display(int[][] arr) {
        for (int i = 0; i < arr.length; i++) {
            for (int j = 0; j < arr.length; j++) {
                System.out.print(arr[i][j] + " ");
            }
            System.out.println();
        }
    }
}

输出:

1 2 3 
4 5 6 
7 8 9 

3 6 9 
2 5 8 
1 4 7 

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;
}

这是c#的

int[,] array = new int[4,4] {
    { 1,2,3,4 },
    { 5,6,7,8 },
    { 9,0,1,2 },
    { 3,4,5,6 }
};

int[,] rotated = RotateMatrix(array, 4);

static int[,] RotateMatrix(int[,] matrix, int n) {
    int[,] ret = new int[n, n];

    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < n; ++j) {
            ret[i, j] = matrix[n - j - 1, i];
        }
    }

    return ret;
}

ruby方式:.transpose。地图&:反向

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;
}

#转置是Ruby的Array类的标准方法，因此:

% irb
irb(main):001:0> m = [[1, 2, 3, 4], [5, 6, 7, 8], [9, 0, 1, 2], [3, 4, 5, 6]]
=> [[1, 2, 3, 4], [5, 6, 7, 8], [9, 0, 1, 2], [3, 4, 5, 6]] 
irb(main):002:0> m.reverse.transpose
=> [[3, 9, 5, 1], [4, 0, 6, 2], [5, 1, 7, 3], [6, 2, 8, 4]]

实现是一个用c写的n^2转置函数，你可以在这里看到: http://www.ruby-doc.org/core-1.9.3/Array.html#method-i-transpose 通过选择“点击切换源”旁边的“转置”。

我记得比O(n^2)的解更好，但只适用于特殊构造的矩阵(如稀疏矩阵)