在Eigen (c++)中:

Eigen::Matrix2d mat;
mat <<  1, 2,
        3, 4;
std::cout << mat << "\n\n";

Eigen::Matrix2d r_plus_90 = mat.transpose().rowwise().reverse();
std::cout << r_plus_90 << "\n\n";

Eigen::Matrix2d r_minus_90 = mat.transpose().colwise().reverse();
std::cout << r_minus_90 << "\n\n";

Eigen::Matrix2d r_180 = mat.colwise().reverse().rowwise().reverse(); // +180 same as -180
std::cout << r_180 << "\n\n";

输出:

1 2
3 4

3 1
4 2

2 4
1 3

4 3
2 1

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

下面是Java语言:

public static void rotateInPlace(int[][] m) {
    for(int layer = 0; layer < m.length/2; layer++){
        int first = layer;
        int last = m.length - 1 - first;
        for(int i = first; i < last; i ++){
            int offset = i - first;
            int top = m[first][i];
            m[first][i] = m[last - offset][first];
            m[last - offset][first] = m[last][last - offset];
            m[last][last - offset] = m[i][last];
            m[i][last] = top;
        }
    }
}

PHP:

<?php    
$a = array(array(1,2,3,4),array(5,6,7,8),array(9,0,1,2),array(3,4,5,6));
$b = array(); //result

while(count($a)>0)
{
    $b[count($a[0])-1][] = array_shift($a[0]);
    if (count($a[0])==0)
    {
         array_shift($a);
    }
}

从PHP5.6开始，数组转位可以通过一个狡猾的array_map()调用来执行。换句话说，列被转换为行。

代码:(演示)

$array = [
    [1, 2, 3, 4],
    [5, 6, 7, 8],
    [9, 0, 1, 2],
    [3, 4, 5, 6]
];
$transposed = array_map(null, ...$array);

美元转置:

[
    [1, 5, 9, 3],
    [2, 6, 0, 4],
    [3, 7, 1, 5],
    [4, 8, 2, 6]
]

这里有大量的好代码，但我只是想以几何形式展示，这样你就能更好地理解代码逻辑。以下是我的处理方法。

首先，不要把这和换位相混淆，换位是很容易的。

基本的想法是把它当作层，我们一次旋转一个层。

假设我们有一辆4x4

1   2   3   4
5   6   7   8
9   10  11  12
13  14  15  16

当我们顺时针旋转90度，我们得到

13  9   5   1
14  10  6   2   
15  11  7   3
16  12  8   4

我们来分解它，首先旋转这四个角

1           4


13          16

然后我们旋转下面这个有点歪斜的菱形

    2
            8
9       
        15

然后是第二个斜菱形

        3
5           
            12
    14

这就搞定了外缘基本上我们一次做一个壳层直到

最后是中间的方块(如果是奇数则是最后一个不动的元素)

6   7
10  11

现在我们来算出每一层的指标，假设我们总是在最外层工作，我们正在做

[0,0] -> [0,n-1], [0,n-1] -> [n-1,n-1], [n-1,n-1] -> [n-1,0], and [n-1,0] -> [0,0]
[0,1] -> [1,n-1], [1,n-2] -> [n-1,n-2], [n-1,n-2] -> [n-2,0], and [n-2,0] -> [0,1]
[0,2] -> [2,n-2], [2,n-2] -> [n-1,n-3], [n-1,n-3] -> [n-3,0], and [n-3,0] -> [0,2]

等等等等 直到我们走到边缘的一半

所以总的来说模式是

[0,i] -> [i,n-i], [i,n-i] -> [n-1,n-(i+1)], [n-1,n-(i+1)] -> [n-(i+1),0], and [n-(i+1),0] to [0,i]

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