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
Python:
rotated = list(zip(*original[::-1]))
和逆时针方向:
rotated_ccw = list(zip(*original))[::-1]
这是如何工作的:
Zip (*original)将通过将列表中的对应项堆叠到新的列表中来交换2d数组的轴。(*操作符告诉函数将包含的列表分布到参数中)
>>> list(zip(*[[1,2,3],[4,5,6],[7,8,9]]))
[[1,4,7],[2,5,8],[3,6,9]]
语句[::-1]反转数组元素(请参阅扩展切片或这个问题):
>>> [[1,2,3],[4,5,6],[7,8,9]][::-1]
[[7,8,9],[4,5,6],[1,2,3]]
最后，将两者结合就得到了旋转变换。
改变[::-1]的位置将使列表在矩阵的不同层次上颠倒。
其他回答
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;
}
}
}
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。
我的c#示例代码的伟大算法发送@dimple:
/* Author: Dudi,
* http://www.tutorialspoint.com/compile_csharp_online.php?PID=0Bw_CjBb95KQMYm5qU3VjVGNuZFU */
using System.IO;
using System;
class Program
{
static void Main()
{
Console.WriteLine("Rotating this matrix by 90+ degree:");
int[,] values=new int[3,3]{{1,2,3}, {4,5,6}, {7,8,9}};
//int[,] values=new int[4,4]{{101,102,103, 104}, {105,106, 107,108}, {109, 110, 111, 112}, {113, 114, 115, 116}};
print2dArray(ref values);
transpose2dArray(ref values);
//print2dArray(ref values);
reverse2dArray(ref values);
Console.WriteLine("Output:");
print2dArray(ref values);
}
static void print2dArray(ref int[,] matrix){
int nLen = matrix.GetLength(0);
int mLen = matrix.GetLength(1);
for(int n=0; n<nLen; n++){
for(int m=0; m<mLen; m++){
Console.Write(matrix[n,m] +"\t");
}
Console.WriteLine();
}
Console.WriteLine();
}
static void transpose2dArray(ref int[,] matrix){
int nLen = matrix.GetLength(0);
int mLen = matrix.GetLength(1);
for(int n=0; n<nLen; n++){
for(int m=0; m<mLen; m++){
if(n>m){
int tmp = matrix[n,m];
matrix[n,m] = matrix[m,n];
matrix[m,n] = tmp;
}
}
}
}
static void reverse2dArray(ref int[,] matrix){
int nLen = matrix.GetLength(0);
int mLen = matrix.GetLength(1);
for(int n=0; n<nLen; n++){
for(int m=0; m<mLen/2; m++){
int tmp = matrix[n,m];
matrix[n,m] = matrix[n, mLen-1-m];
matrix[n,mLen-1-m] = tmp;
}
}
}
}
/*
Rotating this matrix by 90+ degree:
1 2 3
4 5 6
7 8 9
Output:
7 4 1
8 5 2
9 6 3
*/
一些人已经举了一些例子，其中涉及到创建一个新数组。
还有一些需要考虑的事情:
(a)不实际移动数据，只需以不同的方式遍历“旋转”的数组。
(b)就地轮换可能有点棘手。您需要一点空白的地方(大概相当于一行或一列的大小)。有一篇古老的ACM论文是关于进行原地转置的(http://doi.acm.org/10.1145/355719.355729)，但是他们的示例代码是令人讨厌的充满goto的FORTRAN。
附录:
http://doi.acm.org/10.1145/355611.355612是另一种更优越的就地转置算法。
原地旋转不可能比O(n²)更快，原因是如果我们想旋转矩阵，我们必须至少一次触及所有n²元素，无论你实现什么算法。
