如何旋转二维数组?

受Raymond Chen帖子的启发，假设你有一个4x4二维数组，写一个函数使它旋转90度。Raymond链接到伪代码中的解决方案，但我想看到一些真实的东西。

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

就变成:

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

更新:Nick的答案是最直接的，但是有没有比n²更好的方法呢?如果矩阵是10000x10000呢?

当前回答

我只用一个循环就能做到。时间复杂度看起来像O(K)其中K是数组中的所有元素。下面是我用JavaScript做的:

首先，我们用一个数组来表示n^2矩阵。然后，像这样迭代它:

/**
 * Rotates matrix 90 degrees clockwise
 * @param arr: the source array
 * @param n: the array side (array is square n^2)
 */
function rotate (arr, n) {
  var rotated = [], indexes = []

  for (var i = 0; i < arr.length; i++) {
    if (i < n)
      indexes[i] = i * n + (n - 1)
    else
      indexes[i] = indexes[i - n] - 1

    rotated[indexes[i]] = arr[i]
  }
  return rotated
}

基本上，我们转换源数组下标:

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

然后，使用这个转换后的索引数组，我们将实际值放在最终旋转的数组中。

下面是一些测试用例:

//n=3
rotate([
  1, 2, 3,
  4, 5, 6,
  7, 8, 9], 3))

//result:
[7, 4, 1,
 8, 5, 2,
 9, 6, 3]


//n=4
rotate([
  1,  2,  3,  4,
  5,  6,  7,  8,
  9,  10, 11, 12,
  13, 14, 15, 16], 4))

//result:
[13,  9,  5,  1,
 14, 10,  6,  2,
 15, 11,  7,  3,
 16, 12,  8,  4]


//n=5
rotate([
  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], 5))

//result:
[21, 16, 11,  6,  1, 
 22, 17, 12,  7,  2, 
 23, 18, 13,  8,  3, 
 24, 19, 14,  9,  4, 
 25, 20, 15, 10,  5]

2020-05-15 05:15:52

其他回答

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;
}

2008-09-03 20:41:10

Javascript解决NxN矩阵与运行时O(N^2)和内存O(1)

  function rotate90(matrix){
    var length = matrix.length
    for(var row = 0; row < (length / 2); row++){
      for(var col = row; col < ( length - 1 - row); col++){
        var tmpVal = matrix[row][col];
        for(var i = 0; i < 4; i++){
          var rowSwap = col;
          var colSwap = (length - 1) - row;
          var poppedVal = matrix[rowSwap][colSwap];
          matrix[rowSwap][colSwap] = tmpVal;
          tmpVal = poppedVal;
          col = colSwap;
          row = rowSwap;
        }
      }
    }
  }

2015-09-02 03:53:10

为新手程序员，在纯c++。(宝蓝的东西)

#include<iostream.h>
#include<conio.h>

int main()
{
    clrscr();

    int arr[10][10];        // 2d array that holds input elements 
    int result[10][10];     //holds result

    int m,n;                //rows and columns of arr[][]
    int x,y;                //rows and columns of result[][]

    int i,j;                //loop variables
    int t;                  //temporary , holds data while conversion

    cout<<"Enter no. of rows and columns of array: ";
    cin>>m>>n;
    cout<<"\nEnter elements of array: \n\n";
    for(i = 0; i < m; i++)
    {
        for(j = 0; j<n ; j++)
        {
          cin>>arr[i][j];         // input array elements from user
        }
    }


   //rotating matrix by +90 degrees

    x = n ;                      //for non-square matrix
    y = m ;     

    for(i = 0; i < x; i++)
    {  t = m-1;                     // to create required array bounds
       for(j = 0; j < y; j++)
       {
          result[i][j] = arr[t][i];
          t--;
       }
   }

   //print result

   cout<<"\nRotated matrix is: \n\n";
   for(i = 0; i < x; i++)
   {
       for(j = 0; j < y; j++)
       {
             cout<<result[i][j]<<" ";
       }
       cout<<"\n";
   }

   getch();
   return 0;
}

2014-05-06 05:09:02

从线性的角度来看，考虑以下矩阵:

    1 2 3        0 0 1
A = 4 5 6    B = 0 1 0
    7 8 9        1 0 0

现在求A

     1 4 7
A' = 2 5 8
     3 6 9

考虑A'对B的作用，或B对A'的作用。分别为:

      7 4 1          3 6 9
A'B = 8 5 2    BA' = 2 5 8
      9 6 3          1 4 7

这对任何nxn矩阵都是可展开的。在代码中快速应用这个概念:

void swapInSpace(int** mat, int r1, int c1, int r2, int c2)
{
    mat[r1][c1] ^= mat[r2][c2];
    mat[r2][c2] ^= mat[r1][c1];
    mat[r1][c1] ^= mat[r2][c2];
}

void transpose(int** mat, int size)
{
    for (int i = 0; i < size; i++)
    {
        for (int j = (i + 1); j < size; j++)
        {
            swapInSpace(mat, i, j, j, i);
        }
    }
}

void rotate(int** mat, int size)
{
    //Get transpose
    transpose(mat, size);

    //Swap columns
    for (int i = 0; i < size / 2; i++)
    {
        for (int j = 0; j < size; j++)
        {
            swapInSpace(mat, i, j, size - (i + 1), j);
        }
    }
}

2014-10-12 00:47:57