哦，伙计。我一直认为这是一个“我很无聊，我能思考什么”的谜题。我想出了我的原地换位码，但到了这里发现你的和我的几乎一模一样……啊,好。这里是Ruby版本。

require 'pp'
n = 10
a = []
n.times { a << (1..n).to_a }

pp a

0.upto(n/2-1) do |i|
  i.upto(n-i-2) do |j|
    tmp             = 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
  end
end

pp a

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;
}

/* 90-degree clockwise:
   temp_array         = left_col
   left_col           = bottom_row
   bottom_row         = reverse(right_col)
   reverse(right_col) = reverse(top_row)
   reverse(top_row)   = temp_array
*/
void RotateClockwise90(int ** arr, int lo, int hi) {

  if (lo >= hi) 
    return;

  for (int i=lo; i<hi; i++) {
    int j = lo+hi-i;
    int temp   = arr[i][lo];
    arr[i][lo] = arr[hi][i];
    arr[hi][i] = arr[j][hi];
    arr[j][hi] = arr[lo][j];
    arr[lo][j] = temp;
  }

  RotateClockwise90(arr, lo+1, hi-1);
}

c#代码将[n,m] 2D数组向右旋转90度

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace MatrixProject
{
    // mattrix class

    class Matrix{
        private int rows;
        private int cols;
        private int[,] matrix;

        public Matrix(int n){
            this.rows = n;
            this.cols = n;
            this.matrix = new int[this.rows,this.cols];

        }

        public Matrix(int n,int m){
            this.rows = n;
            this.cols = m;

            this.matrix = new int[this.rows,this.cols];
        }

        public void Show()
        {
            for (var i = 0; i < this.rows; i++)
            {
                for (var j = 0; j < this.cols; j++) {
                    Console.Write("{0,3}", this.matrix[i, j]);
                }
                Console.WriteLine();
            }                
        }

        public void ReadElements()
        {
           for (var i = 0; i < this.rows; i++)
                for (var j = 0; j < this.cols; j++)
                {
                    Console.Write("element[{0},{1}]=",i,j);
                    this.matrix[i, j] = Convert.ToInt32(Console.ReadLine());
                }            
        }


        // rotate [n,m] 2D array by 90 deg right
        public void Rotate90DegRight()
        {

            // create a mirror of current matrix
            int[,] mirror = this.matrix;

            // create a new matrix
            this.matrix = new int[this.cols, this.rows];

            for (int i = 0; i < this.rows; i++)
            {
                for (int j = 0; j < this.cols; j++)
                {
                    this.matrix[j, this.rows - i - 1] = mirror[i, j];
                }
            }

            // replace cols count with rows count
            int tmp = this.rows;
            this.rows = this.cols;
            this.cols = tmp;           
        }
    }

    class Program
    {
        static void Main(string[] args)
        {
            Matrix myMatrix = new Matrix(3,4);
            Console.WriteLine("Enter matrix elements:");
            myMatrix.ReadElements();
            Console.WriteLine("Matrix elements are:");
            myMatrix.Show();
            myMatrix.Rotate90DegRight();
            Console.WriteLine("Matrix rotated at 90 deg are:");
            myMatrix.Show();
            Console.ReadLine();
        }
    }
}

结果:

    Enter matrix elements:
    element[0,0]=1
    element[0,1]=2
    element[0,2]=3
    element[0,3]=4
    element[1,0]=5
    element[1,1]=6
    element[1,2]=7
    element[1,3]=8
    element[2,0]=9
    element[2,1]=10
    element[2,2]=11
    element[2,3]=12
    Matrix elements are:
      1  2  3  4
      5  6  7  8
      9 10 11 12
    Matrix rotated at 90 deg are:
      9  5  1
     10  6  2
     11  7  3
     12  8  4

这个解决方案不关心正方形或矩形的尺寸，你可以旋转4x5或5x4甚至4x4，它也不关心大小。 注意，这种实现在每次调用rotate90方法时都会创建一个新数组，它根本不会改变原始数组。

public static void main(String[] args) {
    int[][] a = new int[][] { 
                    { 1, 2, 3, 4 }, 
                    { 5, 6, 7, 8 }, 
                    { 9, 0, 1, 2 }, 
                    { 3, 4, 5, 6 }, 
                    { 7, 8, 9, 0 } 
                  };
    int[][] rotate180 = rotate90(rotate90(a));
    print(rotate180);
}

static int[][] rotate90(int[][] a) {
    int[][] ret = new int[a[0].length][a.length];
    for (int i = 0; i < a.length; i++) {
        for (int j = 0; j < a[i].length; j++) {
            ret[j][a.length - i - 1] = a[i][j];
        }
    }
    return ret;
}

static void print(int[][] array) {
    for (int i = 0; i < array.length; i++) {
        System.out.print("[");
        for (int j = 0; j < array[i].length; j++) {
            System.out.print(array[i][j]);
            System.out.print(" ");
        }
        System.out.println("]");
    }
}

#include <iostream>
#include <iomanip>

using namespace std;
const int SIZE=3;
void print(int a[][SIZE],int);
void rotate(int a[][SIZE],int);

void main()
{
    int a[SIZE][SIZE]={{11,22,33},{44,55,66},{77,88,99}};
    cout<<"the array befor rotate\n";

    print(a,SIZE);
    rotate( a,SIZE);
    cout<<"the array after rotate\n";
    print(a,SIZE);
    cout<<endl;

}

void print(int a[][SIZE],int SIZE)
{
    int i,j;
    for(i=0;i<SIZE;i++)
       for(j=0;j<SIZE;j++)
          cout<<a[i][j]<<setw(4);
}

void rotate(int a[][SIZE],int SIZE)
{
    int temp[3][3],i,j;
    for(i=0;i<SIZE;i++)
       for(j=0;j<SIZE/2.5;j++)
       {
           temp[i][j]= a[i][j];
           a[i][j]= a[j][SIZE-i-1] ;
           a[j][SIZE-i-1] =temp[i][j];

       }
}