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 

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;
}

#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];

       }
}

这是将数组旋转90度的简单C代码。希望这能有所帮助。

#include <stdio.h>

void main(){
int arr[3][4] =     {85, 2, 85,  4,
                     85, 6,  7, 85,
                     9, 85, 11, 12};


int arr1[4][3];

int i = 0, j = 0;

for(i=0;i<4;i++){
int k = 2;//k = (number of columns in the new array arr1 - 1)
for(j=0;j<3;j++){
arr1[i][j]=arr[k][i];
k--;
}
}

int l, m;
for(l=0;l<4;l++){
for(m=0;m<3;m++){
printf("%d ", arr1[l][m]);
}
printf("\n");
}
}//end main

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;
        }
    }
}