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
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;
}
其他回答
这是我在C中的就地实现
void rotateRight(int matrix[][SIZE], int length) {
int layer = 0;
for (int layer = 0; layer < length / 2; ++layer) {
int first = layer;
int last = length - 1 - layer;
for (int i = first; i < last; ++i) {
int topline = matrix[first][i];
int rightcol = matrix[i][last];
int bottomline = matrix[last][length - layer - 1 - i];
int leftcol = matrix[length - layer - 1 - i][first];
matrix[first][i] = leftcol;
matrix[i][last] = topline;
matrix[last][length - layer - 1 - i] = rightcol;
matrix[length - layer - 1 - i][first] = bottomline;
}
}
}
这是一个空间旋转方法，由java编写，只适用于正方形。对于非正方形的2d数组，无论如何都必须创建新数组。
private void rotateInSpace(int[][] arr) {
int z = arr.length;
for (int i = 0; i < z / 2; i++) {
for (int j = 0; j < (z / 2 + z % 2); j++) {
int x = i, y = j;
int temp = arr[x][y];
for (int k = 0; k < 4; k++) {
int temptemp = arr[y][z - x - 1];
arr[y][z - x - 1] = temp;
temp = temptemp;
int tempX = y;
y = z - x - 1;
x = tempX;
}
}
}
}
通过创建新数组旋转任何大小的2d数组的代码:
private int[][] rotate(int[][] arr) {
int width = arr[0].length;
int depth = arr.length;
int[][] re = new int[width][depth];
for (int i = 0; i < depth; i++) {
for (int j = 0; j < width; j++) {
re[j][depth - i - 1] = arr[i][j];
}
}
return re;
}
下面是PHP的递归方法:
$m = array();
$m[0] = array('a', 'b', 'c');
$m[1] = array('d', 'e', 'f');
$m[2] = array('g', 'h', 'i');
$newMatrix = array();
function rotateMatrix($m, $i = 0, &$newMatrix)
{
foreach ($m as $chunk) {
$newChunk[] = $chunk[$i];
}
$newMatrix[] = array_reverse($newChunk);
$i++;
if ($i < count($m)) {
rotateMatrix($m, $i, $newMatrix);
}
}
rotateMatrix($m, 0, $newMatrix);
echo '<pre>';
var_dump($newMatrix);
echo '<pre>';
这是Java中的一个更好的版本:我已经为一个具有不同宽度和高度的矩阵制作了它
H是旋转后矩阵的高度 W是旋转后矩阵的宽度
public int[][] rotateMatrixRight(int[][] matrix)
{
/* W and H are already swapped */
int w = matrix.length;
int h = matrix[0].length;
int[][] ret = new int[h][w];
for (int i = 0; i < h; ++i) {
for (int j = 0; j < w; ++j) {
ret[i][j] = matrix[w - j - 1][i];
}
}
return ret;
}
public int[][] rotateMatrixLeft(int[][] matrix)
{
/* W and H are already swapped */
int w = matrix.length;
int h = matrix[0].length;
int[][] ret = new int[h][w];
for (int i = 0; i < h; ++i) {
for (int j = 0; j < w; ++j) {
ret[i][j] = matrix[j][h - i - 1];
}
}
return ret;
}
此代码基于Nick Berardi的帖子。
c#代码将[n,m] 2D数组向右旋转90度
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
namespace MatrixProject
{
// mattrix class
class Matrix{
private int rows;
private int cols;
private int[,] matrix;
public Matrix(int n){
this.rows = n;
this.cols = n;
this.matrix = new int[this.rows,this.cols];
}
public Matrix(int n,int m){
this.rows = n;
this.cols = m;
this.matrix = new int[this.rows,this.cols];
}
public void Show()
{
for (var i = 0; i < this.rows; i++)
{
for (var j = 0; j < this.cols; j++) {
Console.Write("{0,3}", this.matrix[i, j]);
}
Console.WriteLine();
}
}
public void ReadElements()
{
for (var i = 0; i < this.rows; i++)
for (var j = 0; j < this.cols; j++)
{
Console.Write("element[{0},{1}]=",i,j);
this.matrix[i, j] = Convert.ToInt32(Console.ReadLine());
}
}
// rotate [n,m] 2D array by 90 deg right
public void Rotate90DegRight()
{
// create a mirror of current matrix
int[,] mirror = this.matrix;
// create a new matrix
this.matrix = new int[this.cols, this.rows];
for (int i = 0; i < this.rows; i++)
{
for (int j = 0; j < this.cols; j++)
{
this.matrix[j, this.rows - i - 1] = mirror[i, j];
}
}
// replace cols count with rows count
int tmp = this.rows;
this.rows = this.cols;
this.cols = tmp;
}
}
class Program
{
static void Main(string[] args)
{
Matrix myMatrix = new Matrix(3,4);
Console.WriteLine("Enter matrix elements:");
myMatrix.ReadElements();
Console.WriteLine("Matrix elements are:");
myMatrix.Show();
myMatrix.Rotate90DegRight();
Console.WriteLine("Matrix rotated at 90 deg are:");
myMatrix.Show();
Console.ReadLine();
}
}
}
结果:
Enter matrix elements:
element[0,0]=1
element[0,1]=2
element[0,2]=3
element[0,3]=4
element[1,0]=5
element[1,1]=6
element[1,2]=7
element[1,3]=8
element[2,0]=9
element[2,1]=10
element[2,2]=11
element[2,3]=12
Matrix elements are:
1 2 3 4
5 6 7 8
9 10 11 12
Matrix rotated at 90 deg are:
9 5 1
10 6 2
11 7 3
12 8 4
