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
下面是一个原地旋转的数组，而不是使用一个全新的数组来保存结果。我已经停止了数组的初始化和输出。这只适用于正方形数组，但它们可以是任何大小。内存开销等于数组中一个元素的大小，因此您可以对任意大的数组进行旋转。
int a[4][4];
int n = 4;
int tmp;
for (int i = 0; i < n / 2; i++)
{
for (int j = i; j < n - i - 1; j++)
{
tmp = a[i][j];
a[i][j] = a[j][n-i-1];
a[j][n-i-1] = a[n-i-1][n-j-1];
a[n-i-1][n-j-1] = a[n-j-1][i];
a[n-j-1][i] = tmp;
}
}
其他回答
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。
O(1)内存算法:
旋转最外层的数据，然后你可以得到以下结果: [3] [9] [5] [1] [4] [6] [7] [2] [5] [0] [1] [3] [6] [2] [8] [4]
做这个旋转，我们知道
dest[j][n-1-i] = src[i][j]
观察下图: A (0,0) -> A (0,3) A (0,3) -> A (3,3) A (3,3) -> A (3,0) A (3,0) -> A (0,0)
因此它是一个圆，你可以在一个循环中旋转N个元素。做这个N-1循环，然后你可以旋转最外层的元素。
对于2X2，内部也是一样的问题。
因此，我们可以得出如下结论:
function rotate(array, N)
{
Rotate outer-most data
rotate a new array with N-2 or you can do the similar action following step1
}
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;
}
}
}
PHP:
array_unshift($array, null);
$array = call_user_func_array("array_map", $array);
如果你需要旋转矩形二维阵列90度，在上面的代码之前或之后添加以下一行(取决于你需要的旋转方向):
$array = array_reverse($array);
哦，伙计。我一直认为这是一个“我很无聊，我能思考什么”的谜题。我想出了我的原地换位码，但到了这里发现你的和我的几乎一模一样……啊,好。这里是Ruby版本。
require 'pp'
n = 10
a = []
n.times { a << (1..n).to_a }
pp a
0.upto(n/2-1) do |i|
i.upto(n-i-2) do |j|
tmp = a[i][j]
a[i][j] = a[n-j-1][i]
a[n-j-1][i] = a[n-i-1][n-j-1]
a[n-i-1][n-j-1] = a[j][n-i-1]
a[j][n-i-1] = tmp
end
end
pp a
