在JavaScript中实现dimple的+90伪代码(例如转置然后反转每一行):

function rotate90(a){
  // transpose from http://www.codesuck.com/2012/02/transpose-javascript-array-in-one-line.html
  a = Object.keys(a[0]).map(function (c) { return a.map(function (r) { return r[c]; }); });
  // row reverse
  for (i in a){
    a[i] = a[i].reverse();
  }
  return a;
}

对于i:= 0到X 对于j:= 0到X做 [j][i]:= graphic2[X-i][j]

X是图形所在数组的大小。

基于大量的其他答案，我用c#想出了这个:

/// <param name="rotation">The number of rotations (if negative, the <see cref="Matrix{TValue}"/> is rotated counterclockwise; 
/// otherwise, it's rotated clockwise). A single (positive) rotation is equivalent to 90° or -270°; a single (negative) rotation is 
/// equivalent to -90° or 270°. Matrices may be rotated by 90°, 180°, or 270° only (or multiples thereof).</param>
/// <returns></returns>
public Matrix<TValue> Rotate(int rotation)
{
    var result = default(Matrix<TValue>);

    //This normalizes the requested rotation (for instance, if 10 is specified, the rotation is actually just +-2 or +-180°, but all 
    //correspond to the same rotation).
    var d = rotation.ToDouble() / 4d;
    d = d - (int)d;

    var degree = (d - 1d) * 4d;

    //This gets the type of rotation to make; there are a total of four unique rotations possible (0°, 90°, 180°, and 270°).
    //Each correspond to 0, 1, 2, and 3, respectively (or 0, -1, -2, and -3, if in the other direction). Since
    //1 is equivalent to -3 and so forth, we combine both cases into one. 
    switch (degree)
    {
        case -3:
        case +1:
            degree = 3;
            break;
        case -2:
        case +2:
            degree = 2;
            break;
        case -1:
        case +3:
            degree = 1;
            break;
        case -4:
        case  0:
        case +4:
            degree = 0;
            break;
    }
    switch (degree)
    {
        //The rotation is 0, +-180°
        case 0:
        case 2:
            result = new TValue[Rows, Columns];
            break;
        //The rotation is +-90°
        case 1:
        case 3:
            result = new TValue[Columns, Rows];
            break;
    }

    for (uint i = 0; i < Columns; ++i)
    {
        for (uint j = 0; j < Rows; ++j)
        {
            switch (degree)
            {
                //If rotation is 0°
                case 0:
                    result._values[j][i] = _values[j][i];
                    break;
                //If rotation is -90°
                case 1:
                    //Transpose, then reverse each column OR reverse each row, then transpose
                    result._values[i][j] = _values[j][Columns - i - 1];
                    break;
                //If rotation is +-180°
                case 2:
                    //Reverse each column, then reverse each row
                    result._values[(Rows - 1) - j][(Columns - 1) - i] = _values[j][i];
                    break;
                //If rotation is +90°
                case 3:
                    //Transpose, then reverse each row
                    result._values[i][j] = _values[Rows - j - 1][i];
                    break;
            }
        }
    }
    return result;
}

其中_values对应于由Matrix<TValue>定义的私有二维数组(形式为[][])。result = new TValue[Columns, Rows]可能通过隐式操作符重载并将二维数组转换为Matrix<TValue>。 Columns和Rows两个属性是公共属性，用于获取当前实例的列数和行数:

public uint Columns 
    => (uint)_values[0].Length;

public uint Rows 
    => (uint)_values.Length;

当然，假设您更喜欢使用无符号下标;-)

所有这些都允许您指定它应该旋转多少次，以及它应该向左旋转(如果小于零)还是向右旋转(如果大于零)。您可以改进此方法，以检查实际角度的旋转，但如果值不是90的倍数，则可能会抛出异常。有了这些输入，你可以相应地改变方法:

public Matrix<TValue> Rotate(int rotation)
{
    var _rotation = (double)rotation / 90d;

    if (_rotation - Math.Floor(_rotation) > 0)
    {
        throw new NotSupportedException("A matrix may only be rotated by multiples of 90.").
    }

    rotation = (int)_rotation;
    ...
}

Since a degree is more accurately expressed by double than int, but a matrix can only rotate in multiples of 90, it is far more intuitive to make the argument correspond to something else that can be accurately represented by the data structure used. int is perfect because it can tell you how many times to rotate it up to a certain unit (90) as well as the direction. double may very well be able to tell you that also, but it also includes values that aren't supported by this operation (which is inherently counter-intuitive).

#转置是Ruby的Array类的标准方法，因此:

% irb
irb(main):001:0> m = [[1, 2, 3, 4], [5, 6, 7, 8], [9, 0, 1, 2], [3, 4, 5, 6]]
=> [[1, 2, 3, 4], [5, 6, 7, 8], [9, 0, 1, 2], [3, 4, 5, 6]] 
irb(main):002:0> m.reverse.transpose
=> [[3, 9, 5, 1], [4, 0, 6, 2], [5, 1, 7, 3], [6, 2, 8, 4]]

实现是一个用c写的n^2转置函数，你可以在这里看到: http://www.ruby-doc.org/core-1.9.3/Array.html#method-i-transpose 通过选择“点击切换源”旁边的“转置”。

我记得比O(n^2)的解更好，但只适用于特殊构造的矩阵(如稀疏矩阵)

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;
        }
    }
}

正如我在上一篇文章中所说的，这里有一些c#代码，可以为任何大小的矩阵实现O(1)矩阵旋转。为了简洁性和可读性，没有错误检查或范围检查。代码:

static void Main (string [] args)
{
  int [,]
    //  create an arbitrary matrix
    m = {{0, 1}, {2, 3}, {4, 5}};

  Matrix
    //  create wrappers for the data
    m1 = new Matrix (m),
    m2 = new Matrix (m),
    m3 = new Matrix (m);

  //  rotate the matricies in various ways - all are O(1)
  m1.RotateClockwise90 ();
  m2.Rotate180 ();
  m3.RotateAnitclockwise90 ();

  //  output the result of transforms
  System.Diagnostics.Trace.WriteLine (m1.ToString ());
  System.Diagnostics.Trace.WriteLine (m2.ToString ());
  System.Diagnostics.Trace.WriteLine (m3.ToString ());
}

class Matrix
{
  enum Rotation
  {
    None,
    Clockwise90,
    Clockwise180,
    Clockwise270
  }

  public Matrix (int [,] matrix)
  {
    m_matrix = matrix;
    m_rotation = Rotation.None;
  }

  //  the transformation routines
  public void RotateClockwise90 ()
  {
    m_rotation = (Rotation) (((int) m_rotation + 1) & 3);
  }

  public void Rotate180 ()
  {
    m_rotation = (Rotation) (((int) m_rotation + 2) & 3);
  }

  public void RotateAnitclockwise90 ()
  {
    m_rotation = (Rotation) (((int) m_rotation + 3) & 3);
  }

  //  accessor property to make class look like a two dimensional array
  public int this [int row, int column]
  {
    get
    {
      int
        value = 0;

      switch (m_rotation)
      {
      case Rotation.None:
        value = m_matrix [row, column];
        break;

      case Rotation.Clockwise90:
        value = m_matrix [m_matrix.GetUpperBound (0) - column, row];
        break;

      case Rotation.Clockwise180:
        value = m_matrix [m_matrix.GetUpperBound (0) - row, m_matrix.GetUpperBound (1) - column];
        break;

      case Rotation.Clockwise270:
        value = m_matrix [column, m_matrix.GetUpperBound (1) - row];
        break;
      }

      return value;
    }

    set
    {
      switch (m_rotation)
      {
      case Rotation.None:
        m_matrix [row, column] = value;
        break;

      case Rotation.Clockwise90:
        m_matrix [m_matrix.GetUpperBound (0) - column, row] = value;
        break;

      case Rotation.Clockwise180:
        m_matrix [m_matrix.GetUpperBound (0) - row, m_matrix.GetUpperBound (1) - column] = value;
        break;

      case Rotation.Clockwise270:
        m_matrix [column, m_matrix.GetUpperBound (1) - row] = value;
        break;
      }
    }
  }

  //  creates a string with the matrix values
  public override string ToString ()
  {
    int
      num_rows = 0,
      num_columns = 0;

    switch (m_rotation)
    {
    case Rotation.None:
    case Rotation.Clockwise180:
      num_rows = m_matrix.GetUpperBound (0);
      num_columns = m_matrix.GetUpperBound (1);
      break;

    case Rotation.Clockwise90:
    case Rotation.Clockwise270:
      num_rows = m_matrix.GetUpperBound (1);
      num_columns = m_matrix.GetUpperBound (0);
      break;
    }

    StringBuilder
      output = new StringBuilder ();

    output.Append ("{");

    for (int row = 0 ; row <= num_rows ; ++row)
    {
      if (row != 0)
      {
        output.Append (", ");
      }

      output.Append ("{");

      for (int column = 0 ; column <= num_columns ; ++column)
      {
        if (column != 0)
        {
          output.Append (", ");
        }

        output.Append (this [row, column].ToString ());
      }

      output.Append ("}");
    }

    output.Append ("}");

    return output.ToString ();
  }

  int [,]
    //  the original matrix
    m_matrix;

  Rotation
    //  the current view of the matrix
    m_rotation;
}

好的，我把手举起来，当旋转时，它实际上不会对原始数组做任何修改。但是，在面向对象系统中，只要对象看起来像是被旋转到类的客户端，这就无关紧要了。目前，Matrix类使用对原始数组数据的引用，因此改变m1的任何值也将改变m2和m3。对构造函数稍加更改，创建一个新数组并将值复制到该数组中，就可以将其整理出来。