public static void rotateMatrix(int[,] matrix)
    {
        //C#, to rotate an N*N matrix in place
        int n = matrix.GetLength(0);
        int layers =  n / 2;
        int temp, temp2;

        for (int i = 0; i < layers; i++) // for a 5 * 5 matrix, layers will be 2, since at layer three there would be only one element, (2,2), and we do not need to rotate it with itself 
        {
            int offset = 0;
            while (offset < n - 2 * i - 1)
            {
                // top right <- top left 
                temp = matrix[i + offset, n - i - 1]; //top right value when offset is zero
                matrix[i + offset, n - i - 1] = matrix[i, i + offset];   

                //bottom right <- top right 
                temp2 = matrix[n - i - 1, n - i - 1 - offset]; //bottom right value when offset is zero
                matrix[n - i - 1, n - i - 1 - offset] = temp;  

                //bottom left <- bottom right 
                temp = matrix[n - i - 1 - offset, i];
                matrix[n - i - 1 - offset, i] = temp2;  

                //top left <- bottom left 
                matrix[i, i + offset] = temp; 

                offset++;
            }
        }
    }

这是我的实现，在C, O(1)内存复杂度，原地旋转，顺时针90度:

#include <stdio.h>

#define M_SIZE 5

static void initMatrix();
static void printMatrix();
static void rotateMatrix();

static int m[M_SIZE][M_SIZE];

int main(void){
    initMatrix();
    printMatrix();
    rotateMatrix();
    printMatrix();

    return 0;
}

static void initMatrix(){
    int i, j;

    for(i = 0; i < M_SIZE; i++){
        for(j = 0; j < M_SIZE; j++){
            m[i][j] = M_SIZE*i + j + 1;
        }
    }
}

static void printMatrix(){
    int i, j;

    printf("Matrix\n");
    for(i = 0; i < M_SIZE; i++){
        for(j = 0; j < M_SIZE; j++){
            printf("%02d ", m[i][j]);
        }
        printf("\n");
    }
    printf("\n");
}

static void rotateMatrix(){
    int r, c;

    for(r = 0; r < M_SIZE/2; r++){
        for(c = r; c < M_SIZE - r - 1; c++){
            int tmp = m[r][c];

            m[r][c] = m[M_SIZE - c - 1][r];
            m[M_SIZE - c - 1][r] = m[M_SIZE - r - 1][M_SIZE - c - 1];
            m[M_SIZE - r - 1][M_SIZE - c - 1] = m[c][M_SIZE - r - 1];
            m[c][M_SIZE - r - 1] = tmp;
        }
    }
}

ruby方式:.transpose。地图&:反向

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;
}

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]
]

时间- O(N)，空间- O(1)

public void rotate(int[][] matrix) {
    int n = matrix.length;
    for (int i = 0; i < n / 2; i++) {
        int last = n - 1 - i;
        for (int j = i; j < last; j++) {
            int top = matrix[i][j];
            matrix[i][j] = matrix[last - j][i];
            matrix[last - j][i] = matrix[last][last - j];
            matrix[last][last - j] = matrix[j][last];
            matrix[j][last] = top;
        }
    }
}