这是我对矩阵90度旋转的尝试，这是c中的2步解决方案，首先转置矩阵，然后交换cols。

#define ROWS        5
#define COLS        5

void print_matrix_b(int B[][COLS], int rows, int cols) 
{
    for (int i = 0; i <= rows; i++) {
        for (int j = 0; j <=cols; j++) {
            printf("%d ", B[i][j]);
        }
        printf("\n");
    }
}

void swap_columns(int B[][COLS], int l, int r, int rows)
{
    int tmp;
    for (int i = 0; i <= rows; i++) {
        tmp = B[i][l];
        B[i][l] = B[i][r];
        B[i][r] = tmp;
    }
}


void matrix_2d_rotation(int B[][COLS], int rows, int cols)
{
    int tmp;
    // Transpose the matrix first
    for (int i = 0; i <= rows; i++) {
        for (int j = i; j <=cols; j++) {
            tmp = B[i][j];
            B[i][j] = B[j][i];
            B[j][i] = tmp;
        }
    }
    // Swap the first and last col and continue until
    // the middle.
    for (int i = 0; i < (cols / 2); i++)
        swap_columns(B, i, cols - i, rows);
}



int _tmain(int argc, _TCHAR* argv[])
{
    int B[ROWS][COLS] = { 
                  {1, 2, 3, 4, 5}, 
                      {6, 7, 8, 9, 10},
                          {11, 12, 13, 14, 15},
                          {16, 17, 18, 19, 20},
                          {21, 22, 23, 24, 25}
                        };

    matrix_2d_rotation(B, ROWS - 1, COLS - 1);

    print_matrix_b(B, ROWS - 1, COLS -1);
    return 0;
}

JavaScript解决方案旋转矩阵90度的地方:

function rotateBy90(m) {
  var length = m.length;
  //for each layer of the matrix
  for (var first = 0; first < length >> 1; first++) {
    var last = length - 1 - first;
    for (var i = first; i < last; i++) {
      var top = m[first][i]; //store top
      m[first][i] = m[last - i][first]; //top = left
      m[last - i][first] = m[last][last - i]; //left = bottom
      m[last][last - i] = m[i][last]; //bottom = right
      m[i][last] = top; //right = top
    }
  }
  return m;
}

基于社区wiki算法和这个转置数组的SO答案，这里是一个Swift 4版本，可以逆时针旋转一些2D数组90度。这里假设matrix是一个2D数组:

func rotate(matrix: [[Int]]) -> [[Int]] {
    let transposedPoints = transpose(input: matrix)
    let rotatedPoints = transposedPoints.map{ Array($0.reversed()) }
    return rotatedPoints
}


fileprivate func transpose<T>(input: [[T]]) -> [[T]] {
    if input.isEmpty { return [[T]]() }
    let count = input[0].count
    var out = [[T]](repeating: [T](), count: count)
    for outer in input {
        for (index, inner) in outer.enumerated() {
            out[index].append(inner)
        }
    }

    return out
}

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]
]

这是一个Javascript解决方案:

const transpose = m => m[0].map((x,i) => m.map(x => x[i]));

a: // original matrix
123
456
789

transpose(a).reverse(); // rotate 90 degrees counter clockwise 
369
258
147

transpose(a.slice().reverse()); // rotate 90 degrees clockwise 
741
852
963

transpose(transpose(a.slice().reverse()).slice().reverse())
// rotate 180 degrees 
987
654
321

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("]");
    }
}