受Raymond Chen's post的启发，假设您有一个4x4二维数组，请编写一个将其旋转90度的函数。 Raymond链接到伪代码的解决方案，但我希望看到一些现实世界的东西。
[1][2][3][4]
[5][6][7][8]
[9][0][1][2]
[3][4][5][6]
变为：
[3][9][5][1]
[4][0][6][2]
[5][1][7][3]
[6][2][8][4]
更新：尼克的回答是最简单的，但有没有办法比n ^ 2更好？如果矩阵是10000x10000怎么办？
答案 0 :(得分：368)
O（n ^ 2）时间和O（1）空间算法（没有任何变通办法和手帕的东西！）
旋转+90：
按-90：
旋转方法1：
方法2：
旋转+180：
方法1 ：旋转+90两次
方法2 ：反转每一行然后反转每一列（Transpose）
旋转-180：
方法1 ：旋转-90两次
方法2 ：反转每一列，然后反转每一行
方法3 ：旋转+180，因为它们相同
答案 1 :(得分：137)
这是在C＃
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;
}
答案 2 :(得分：124)
<强>的Python：
rotated = zip(*original[::-1]) # On Python 3, list(zip(*original[::-1]))
便宜，我知道。
逆时针：
rotated_ccw = zip(*original)[::-1] # On Python 3, list(zip(*original))[::-1]
这是如何运作的：（在评论中要求）
zip(*original)将通过将列表中的相应项目堆叠到新列表中来交换2d阵列的轴。 （*运算符告诉函数将包含的列表分配到参数中）
>>> zip(*[[1,2,3],[4,5,6],[7,8,9]])
[[1,4,7],[2,5,8],[3,6,9]]
[::-1]语句会反转数组元素（请参阅Extended Slices）。
>>> [[1,2,3],[4,5,6],[7,8,9]][::-1]
[[7,8,9],[4,5,6],[1,2,3]]
最后，将两者结合将导致旋转变换。
[::-1]的展示位置变化会反转矩阵不同级别的列表。
答案 3 :(得分：68)
这是一个进行旋转而不是使用全新数组来保存结果的旋转。我已经停止了阵列的初始化并将其打印出来。这仅适用于方形阵列，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，因此您可以根据需要旋转大小的数组。
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;
}
}
答案 4 :(得分：38)
这里有很多好的代码，但我只想展示几何上发生的事情，这样你就可以更好地理解代码逻辑了。以下是我将如何做到这一点。
首先，不要将此与转置混淆，这很容易..
basica的想法是将其视为图层，我们一次旋转一层..
说我们有4x4
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
顺时针旋转90后我们得到
13 9 5 1
14 10 6 2
15 11 7 3
16 12 8 4
所以让我们分解一下，首先我们基本上旋转4个角落
1 4
13 16
然后我们旋转下面的钻石，这是一种歪斜的
2
8
9
15
然后是第二颗歪斜的钻石
3
5
12
14
所以这样照顾外边缘，所以我们一次只做一个壳直到
最后是中间的方格（或者如果奇怪的是最后一个元素不移动）
6 7
10 11
所以现在让我们弄清楚每一层的索引，假设我们总是使用最外层，我们正在做
[0,0] -> [0,n-1], [0,n-1] -> [n-1,n-1], [n-1,n-1] -> [n-1,0], and [n-1,0] -> [0,0]
[0,1] -> [1,n-1], [1,n-2] -> [n-1,n-2], [n-1,n-2] -> [n-2,0], and [n-2,0] -> [0,1]
[0,2] -> [2,n-2], [2,n-2] -> [n-1,n-3], [n-1,n-3] -> [n-3,0], and [n-3,0] -> [0,2]
所以一般来说模式是
[0,i] -> [i,n-i], [i,n-i] -> [n-1,n-(i+1)], [n-1,n-(i+1)] -> [n-(i+1),0], and [n-(i+1),0] to [0,i]
答案 5 :(得分：35)
正如我在上一篇文章中所说的，这里是C＃中的一些代码，它为任何大小的矩阵实现O（1）矩阵旋转。为了简洁和可读性，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，我会举起手，旋转时它实际上并没有对原始阵列进行任何修改。但是，在OO系统中，只要对象看起来像是旋转到类的客户端那么无关紧要。目前，Matrix类使用对原始数组数据的引用，因此更改m1的任何值也将更改m2和m3。对构造函数进行一些小改动以创建一个新数组并将值复制到它将对其进行排序。
答案 6 :(得分：23)
虽然可能需要轮换数据（可能是为了更新物理存储的表示），但是在数组访问中添加一个间接层（可能是一个接口）变得更简单并且可能更高效：
interface IReadableMatrix
{
int GetValue(int x, int y);
}
如果您的Matrix已经实现了此界面，则可以通过decorator类进行轮换，如下所示：
class RotatedMatrix : IReadableMatrix
{
private readonly IReadableMatrix _baseMatrix;
public RotatedMatrix(IReadableMatrix baseMatrix)
{
_baseMatrix = baseMatrix;
}
int GetValue(int x, int y)
{
// transpose x and y dimensions
return _baseMatrix(y, x);
}
}
旋转+ 90 / -90 / 180度，水平/垂直翻转和缩放都可以这种方式实现。
需要在特定情况下衡量绩效。但是，O（n ^ 2）操作现在已被O（1）调用替换。这是一个虚拟方法调用， 比直接数组访问慢，所以它取决于旋转后旋转数组的使用频率。如果它被使用一次，那么这种方法肯定会赢。如果它被旋转然后在长时间运行的系统中使用了几天，那么就地旋转可能会表现得更好。这还取决于您是否可以接受前期费用。
与所有绩效问题一样，衡量，衡量，衡量！
答案 7 :(得分：17)
这是Java中的一个更好的版本：我已经为具有不同宽度和高度的矩阵制作了它
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。
答案 8 :(得分：17)
Ruby-way：.transpose.map &:reverse
答案 9 :(得分：13)
已经有很多答案，我发现有两个声称O（1）时间复杂度。 真实 O（1）算法是保持数组存储不变，并更改索引元素的方式。这里的目标是它不消耗额外的内存，也不需要额外的时间来迭代数据。
90度，-90度和180度的旋转是简单的变换，只要您知道2D阵列中有多少行和列，就可以执行这些变换。要将任何矢量旋转90度，请交换轴并取消Y轴。对于-90度，交换轴并取消X轴。对于180度，在没有交换的情况下否定两个轴。
可以进行进一步的转换，例如通过独立地取消轴来水平和/或垂直镜像。
这可以通过例如一种存取方法。以下示例是JavaScript函数，但这些概念同样适用于所有语言。
// Get an array element in column/row order
var getArray2d = function(a, x, y) {
return a[y][x];
};
//demo
var arr = [
[5, 4, 6],
[1, 7, 9],
[-2, 11, 0],
[8, 21, -3],
[3, -1, 2]
];
var newarr = [];
arr[0].forEach(() => newarr.push(new Array(arr.length)));
for (var i = 0; i < newarr.length; i++) {
for (var j = 0; j < newarr[0].length; j++) {
newarr[i][j] = getArray2d(arr, i, j);
}
}
console.log(newarr);
// Get an array element rotated 90 degrees clockwise
function getArray2dCW(a, x, y) {
var t = x;
x = y;
y = a.length - t - 1;
return a[y][x];
}
//demo
var arr = [
[5, 4, 6],
[1, 7, 9],
[-2, 11, 0],
[8, 21, -3],
[3, -1, 2]
];
var newarr = [];
arr[0].forEach(() => newarr.push(new Array(arr.length)));
for (var i = 0; i < newarr[0].length; i++) {
for (var j = 0; j < newarr.length; j++) {
newarr[j][i] = getArray2dCW(arr, i, j);
}
}
console.log(newarr);
// Get an array element rotated 90 degrees counter-clockwise
function getArray2dCCW(a, x, y) {
var t = x;
x = a[0].length - y - 1;
y = t;
return a[y][x];
}
//demo
var arr = [
[5, 4, 6],
[1, 7, 9],
[-2, 11, 0],
[8, 21, -3],
[3, -1, 2]
];
var newarr = [];
arr[0].forEach(() => newarr.push(new Array(arr.length)));
for (var i = 0; i < newarr[0].length; i++) {
for (var j = 0; j < newarr.length; j++) {
newarr[j][i] = getArray2dCCW(arr, i, j);
}
}
console.log(newarr);
// Get an array element rotated 180 degrees
function getArray2d180(a, x, y) {
x = a[0].length - x - 1;
y = a.length - y - 1;
return a[y][x];
}
//demo
var arr = [
[5, 4, 6],
[1, 7, 9],
[-2, 11, 0],
[8, 21, -3],
[3, -1, 2]
];
var newarr = [];
arr.forEach(() => newarr.push(new Array(arr[0].length)));
for (var i = 0; i < newarr[0].length; i++) {
for (var j = 0; j < newarr.length; j++) {
newarr[j][i] = getArray2d180(arr, i, j);
}
}
console.log(newarr);
此代码假定嵌套数组的数组，其中每个内部数组都是一行。
该方法允许您读取（或写入）元素（即使是以随机顺序），就像数组已被旋转或转换一样。现在只需选择正确的函数来调用，可能是通过引用，然后离开你去！
该概念可以扩展为通过访问器方法加性地（和非破坏性地）应用变换。包括任意角度旋转和缩放。
答案 10 :(得分：10)
有几个人已经提出了涉及制作新阵列的例子。
要考虑的其他一些事项：
（a）不是实际移动数据，而是简单地以不同方式遍历“旋转”数组。
（b）就地进行旋转可能有点棘手。你需要一些划痕的地方（大概相当于一行或一列的大小）。有一篇古老的ACM论文关于进行就地转置（http://doi.acm.org/10.1145/355719.355729），但他们的示例代码是讨厌的转载FORTRAN。
附录：
http://doi.acm.org/10.1145/355611.355612是另一种据称优越的就地转置算法。
答案 11 :(得分：8)
Nick's回答也可以用于NxM数组，只需要很小的修改（而不是NxN）。
string[,] orig = new string[n, m];
string[,] rot = new string[m, n];
...
for ( int i=0; i < n; i++ )
for ( int j=0; j < m; j++ )
rot[j, n - i - 1] = orig[i, j];
考虑这个问题的一种方法是将轴的中心（0,0）从左上角移动到右上角。你只是简单地从一个转移到另一个。
答案 12 :(得分：6)
时间 - O（N），空格 - 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;
}
}
}
答案 13 :(得分：5)
这是我的Ruby版本（请注意，值不会显示相同，但仍按所述方式旋转）。
def rotate(matrix)
result = []
4.times { |x|
result[x] = []
4.times { |y|
result[x][y] = matrix[y][3 - x]
}
}
result
end
matrix = []
matrix[0] = [1,2,3,4]
matrix[1] = [5,6,7,8]
matrix[2] = [9,0,1,2]
matrix[3] = [3,4,5,6]
def print_matrix(matrix)
4.times { |y|
4.times { |x|
print "#{matrix[x][y]} "
}
puts ""
}
end
print_matrix(matrix)
puts ""
print_matrix(rotate(matrix))
输出：
1 5 9 3
2 6 0 4
3 7 1 5
4 8 2 6
4 3 2 1
8 7 6 5
2 1 0 9
6 5 4 3
答案 14 :(得分：4)
这是一个空间旋转方法，由java，仅用于square。对于非方形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;
}
}
}
}
通过创建新数组来旋转任何大小的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;
}
答案 15 :(得分：3)
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;
}
答案 16 :(得分：3)
您可以通过 3个简单步骤：
执行此操作1 ）假设我们有一个矩阵
1 2 3
4 5 6
7 8 9
2 ）进行矩阵的转置
1 4 7
2 5 8
3 6 9
3 ）交换行以获得旋转矩阵
3 6 9
2 5 8
1 4 7
Java 源代码：
public class MyClass {
public static void main(String args[]) {
Demo obj = new Demo();
/*initial matrix to rotate*/
int[][] matrix = { { 1, 2, 3 }, { 4, 5, 6 }, { 7, 8, 9 } };
int[][] transpose = new int[3][3]; // matrix to store transpose
obj.display(matrix); // initial matrix
obj.rotate(matrix, transpose); // call rotate method
System.out.println();
obj.display(transpose); // display the rotated matix
}
}
class Demo {
public void rotate(int[][] mat, int[][] tran) {
/* First take the transpose of the matrix */
for (int i = 0; i < mat.length; i++) {
for (int j = 0; j < mat.length; j++) {
tran[i][j] = mat[j][i];
}
}
/*
* Interchange the rows of the transpose matrix to get rotated
* matrix
*/
for (int i = 0, j = tran.length - 1; i != j; i++, j--) {
for (int k = 0; k < tran.length; k++) {
swap(i, k, j, k, tran);
}
}
}
public void swap(int a, int b, int c, int d, int[][] arr) {
int temp = arr[a][b];
arr[a][b] = arr[c][d];
arr[c][d] = temp;
}
/* Method to display the matrix */
public void display(int[][] arr) {
for (int i = 0; i < arr.length; i++) {
for (int j = 0; j < arr.length; j++) {
System.out.print(arr[i][j] + " ");
}
System.out.println();
}
}
}
<强>输出：
1 2 3
4 5 6
7 8 9
3 6 9
2 5 8
1 4 7
答案 17 :(得分：2)
这是我的实现，在C，O（1）内存复杂度，就地旋转，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;
}
}
}
答案 18 :(得分：2)
C＃代码将[n，m] 2D阵列向右旋转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
答案 19 :(得分：2)
PHP：
<?php
$a = array(array(1,2,3,4),array(5,6,7,8),array(9,0,1,2),array(3,4,5,6));
$b = array(); //result
while(count($a)>0)
{
$b[count($a[0])-1][] = array_shift($a[0]);
if (count($a[0])==0)
{
array_shift($a);
}
}
?>
答案 20 :(得分：2)
从线性的角度来看，考虑矩阵：
1 2 3 0 0 1
A = 4 5 6 B = 0 1 0
7 8 9 1 0 0
现在进行A转置
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
对于任何n x n矩阵，这都是可扩展的。 并在代码中快速应用此概念：
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);
}
}
}
答案 21 :(得分：2)
以下是Java版本：
public static void rightRotate(int[][] matrix, int n) {
for (int layer = 0; layer < n / 2; layer++) {
int first = layer;
int last = n - 1 - first;
for (int i = first; i < last; i++) {
int offset = i - first;
int temp = matrix[first][i];
matrix[first][i] = matrix[last-offset][first];
matrix[last-offset][first] = matrix[last][last-offset];
matrix[last][last-offset] = matrix[i][last];
matrix[i][last] = temp;
}
}
}
该方法首先旋转最外层，然后顺序移动到内层。
答案 22 :(得分：1)
NxN矩阵的Javascript解决方案，具有运行时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;
}
}
}
}
答案 23 :(得分：1)
很棒的答案，但对于那些正在为此寻找干燥JavaScript代码的人 - +90度和-90度：
// Input: 1 2 3
// 4 5 6
// 7 8 9
// Transpose:
// 1 4 7
// 2 5 8
// 3 6 9
// Output:
// +90 Degree:
// 7 4 1
// 8 5 2
// 9 6 3
// -90 Degree:
// 3 6 9
// 2 5 8
// 1 4 7
// Rotate +90
function rotate90(matrix) {
matrix = transpose(matrix);
matrix.map(function(array) {
array.reverse();
});
return matrix;
}
// Rotate -90
function counterRotate90(matrix) {
var result = createEmptyMatrix(matrix.length);
matrix = transpose(matrix);
var counter = 0;
for (var i = matrix.length - 1; i >= 0; i--) {
result[counter] = matrix[i];
counter++;
}
return result;
}
// Create empty matrix
function createEmptyMatrix(len) {
var result = new Array();
for (var i = 0; i < len; i++) {
result.push([]);
}
return result;
}
// Transpose the matrix
function transpose(matrix) {
// make empty array
var len = matrix.length;
var result = createEmptyMatrix(len);
for (var i = 0; i < matrix.length; i++) {
for (var j = 0; j < matrix[i].length; j++) {
var temp = matrix[i][j];
result[j][i] = temp;
}
}
return result;
}
// Test Cases
var array1 = [
[1, 2],
[3, 4]
];
var array2 = [
[1, 2, 3],
[4, 5, 6],
[7, 8, 9]
];
var array3 = [
[1, 2, 3, 4],
[5, 6, 7, 8],
[9, 10, 11, 12],
[13, 14, 15, 16]
];
// +90 degress Rotation Tests
var test1 = rotate90(array1);
var test2 = rotate90(array2);
var test3 = rotate90(array3);
console.log(test1);
console.log(test2);
console.log(test3);
// -90 degress Rotation Tests
var test1 = counterRotate90(array1);
var test2 = counterRotate90(array2);
var test3 = counterRotate90(array3);
console.log(test1);
console.log(test2);
console.log(test3);
答案 24 :(得分：1)
PHP解决方案适用于顺时针和＆amp;逆时针
$aMatrix = array(
array( 1, 2, 3 ),
array( 4, 5, 6 ),
array( 7, 8, 9 )
);
function CounterClockwise( $aMatrix )
{
$iCount = count( $aMatrix );
$aReturn = array();
for( $y = 0; $y < $iCount; ++$y )
{
for( $x = 0; $x < $iCount; ++$x )
{
$aReturn[ $iCount - $x - 1 ][ $y ] = $aMatrix[ $y ][ $x ];
}
}
return $aReturn;
}
function Clockwise( $aMatrix )
{
$iCount = count( $aMatrix );
$aReturn = array();
for( $y = 0; $y < $iCount; ++$y )
{
for( $x = 0; $x < $iCount; ++$x )
{
$aReturn[ $x ][ $iCount - $y - 1 ] = $aMatrix[ $y ][ $x ];
}
}
return $aReturn;
}
function printMatrix( $aMatrix )
{
$iCount = count( $aMatrix );
for( $x = 0; $x < $iCount; ++$x )
{
for( $y = 0; $y < $iCount; ++$y )
{
echo $aMatrix[ $x ][ $y ];
echo " ";
}
echo "\n";
}
}
printMatrix( $aMatrix );
echo "\n";
$aNewMatrix = CounterClockwise( $aMatrix );
printMatrix( $aNewMatrix );
echo "\n";
$aNewMatrix = Clockwise( $aMatrix );
printMatrix( $aNewMatrix );
答案 25 :(得分：1)
在蟒蛇中：
print(bool([x for x in input().split() if int(x) > 0 and x==x[::-1]]))
答案 26 :(得分：1)
private static int[][] rotate(int[][] matrix, int n) {
int[][] rotated = new int[n][n];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
rotated[i][j] = matrix[n-j-1][i];
}
}
return rotated;
}
答案 27 :(得分：1)
@dagorym：噢，伙计。我一直把它当成一个好“我很无聊，我能思考什么”这个难题。我提出了我的就地换位代码，但到了这里找到你的几乎和我的相同......啊，好吧。这是在Ruby中。
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
答案 28 :(得分：1)
short normal[4][4] = {{8,4,7,5},{3,4,5,7},{9,5,5,6},{3,3,3,3}};
short rotated[4][4];
for (int r = 0; r < 4; ++r)
{
for (int c = 0; c < 4; ++c)
{
rotated[r][c] = normal[c][3-r];
}
}
简单的C ++方法，在大数组中会有很大的内存开销。
答案 29 :(得分：1)
For i:= 0 to X do
For j := 0 to X do
graphic[j][i] := graphic2[X-i][j]
X是图形所在数组的大小。
答案 30 :(得分：1)
#transpose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]]
实现是用C编写的n ^ 2转置函数。你可以在这里看到它： http://www.ruby-doc.org/core-1.9.3/Array.html#method-i-transpose 选择“转置”旁边的“点击切换源”。
我记得比O（n ^ 2）解决方案更好，但仅适用于特殊构造的矩阵（例如稀疏矩阵）
答案 31 :(得分：1)
对于任何M * N矩阵，矩阵旋转90度的C代码顺时针IN PLACE
void rotateInPlace(int * arr[size][size], int row, int column){
int i, j;
int temp = row>column?row:column;
int flipTill = row < column ? row : column;
for(i=0;i<flipTill;i++){
for(j=0;j<i;j++){
swapArrayElements(arr, i, j);
}
}
temp = j+1;
for(i = row>column?i:0; i<row; i++){
for(j=row<column?temp:0; j<column; j++){
swapArrayElements(arr, i, j);
}
}
for(i=0;i<column;i++){
for(j=0;j<row/2;j++){
temp = arr[i][j];
arr[i][j] = arr[i][row-j-1];
arr[i][row-j-1] = temp;
}
}
}
答案 32 :(得分：1)
这是我对矩阵90度旋转的尝试，这是C中的两步解决方案。首先将矩阵转置到位，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;
}
答案 33 :(得分：1)
这是我在C中的实施实施
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;
}
}
}
答案 34 :(得分：0)
顺时针或逆时针旋转2D数组的常用方法。
1 2 3 7 8 9 7 4 1
4 5 6 => 4 5 6 => 8 5 2
7 8 9 1 2 3 9 6 3
void rotate(vector<vector<int> > &matrix) {
reverse(matrix.begin(), matrix.end());
for (int i = 0; i < matrix.size(); ++i) {
for (int j = i + 1; j < matrix[i].size(); ++j)
swap(matrix[i][j], matrix[j][i]);
}
}
1 2 3 3 2 1 3 6 9
4 5 6 => 6 5 4 => 2 5 8
7 8 9 9 8 7 1 4 7
void anti_rotate(vector<vector<int> > &matrix) {
for (auto vi : matrix) reverse(vi.begin(), vi.end());
for (int i = 0; i < matrix.size(); ++i) {
for (int j = i + 1; j < matrix[i].size(); ++j)
swap(matrix[i][j], matrix[j][i]);
}
}
答案 35 :(得分：0)
我的轮换版本：
void rotate_matrix(int *matrix, int size)
{
int result[size*size];
for (int i = 0; i < size; ++i)
for (int j = 0; j < size; ++j)
result[(size - 1 - i) + j*size] = matrix[i*size+j];
for (int i = 0; i < size*size; ++i)
matrix[i] = result[i];
}
在其中我们将最后一列更改为第一行，因此更进一步。它可能不是最佳的，但理解清晰。
答案 36 :(得分：0)
在Eigen（C ++）中：
Eigen::Matrix2d mat;
mat << 1, 2,
3, 4;
std::cout << mat << "\n\n";
Eigen::Matrix2d r_plus_90 = mat.transpose().rowwise().reverse();
std::cout << r_plus_90 << "\n\n";
Eigen::Matrix2d r_minus_90 = mat.transpose().colwise().reverse();
std::cout << r_minus_90 << "\n\n";
Eigen::Matrix2d r_180 = mat.colwise().reverse().rowwise().reverse(); // +180 same as -180
std::cout << r_180 << "\n\n";
输出：
1 2
3 4
3 1
4 2
2 4
1 3
4 3
2 1
答案 37 :(得分：0)
这是一种递归的PHP方式：
$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>';
答案 38 :(得分：0)
我可以通过单循环来做到这一点。时间复杂度看起来像 O（K），其中K是数组的所有项目。 这是我在JavaScript中完成的方法：
首先，我们用单个数组表示n ^ 2矩阵。然后，像这样遍历它：
/**
* 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，我们转换源数组索引：
[0,1,2,3,4,5,6,7,8] => [2,5,8,1,4,7,0,3,6]
然后，使用此转换后的indexes数组，将实际值放置在最后的rotated数组中。
以下是一些测试用例：
//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]
答案 39 :(得分：0)
此解决方案不在乎正方形或矩形的尺寸，您可以旋转4x5或5x4甚至4x4，它也不在乎尺寸。 请注意，此实现每次调用Rotate90方法都会创建一个新数组，它根本不会改变原始数组。
public static void main(String[] args) {
int[][] a = new int[][] {
{ 1, 2, 3, 4 },
{ 5, 6, 7, 8 },
{ 9, 0, 1, 2 },
{ 3, 4, 5, 6 },
{ 7, 8, 9, 0 }
};
int[][] rotate180 = rotate90(rotate90(a));
print(rotate180);
}
static int[][] rotate90(int[][] a) {
int[][] ret = new int[a[0].length][a.length];
for (int i = 0; i < a.length; i++) {
for (int j = 0; j < a[i].length; j++) {
ret[j][a.length - i - 1] = a[i][j];
}
}
return ret;
}
static void print(int[][] array) {
for (int i = 0; i < array.length; i++) {
System.out.print("[");
for (int j = 0; j < array[i].length; j++) {
System.out.print(array[i][j]);
System.out.print(" ");
}
System.out.println("]");
}
}
答案 40 :(得分：0)
基于社区维基算法和this SO answer用于转置数组，这里有一个Swift 4版本可以逆时针旋转一些2D数组90度。这假定matrix是一个2D数组：
func rotate(matrix: [[Int]]) -> [[Int]] {
let transposedPoints = transpose(input: matrix)
let rotatedPoints = transposedPoints.map{ Array($0.reversed()) }
return rotatedPoints
}
fileprivate func transpose<T>(input: [[T]]) -> [[T]] {
if input.isEmpty { return [[T]]() }
let count = input[0].count
var out = [[T]](repeating: [T](), count: count)
for outer in input {
for (index, inner) in outer.enumerated() {
out[index].append(inner)
}
}
return out
}
答案 41 :(得分：0)
根据过多的其他答案，我在C＃中提出了这个问题：
/// <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;
}
_values对应于由Matrix<TValue>定义的私有二维数组（以[][]的形式）。 result = new TValue[Columns, Rows]可以通过隐式运算符重载实现，并将二维数组转换为Matrix<TValue>。
两个属性Columns和Rows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索引; - ）
所有这些允许您指定应旋转的次数以及是否应向左旋转（如果小于零）或向右旋转（如果大于零）。您可以改进这一点以检查实际度数的旋转，但是如果值不是90的倍数，则您想要抛出异常。使用该输入，您可以相应地更改方法：
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;
...
}
由于double比int更准确地表达了一个度，但是一个矩阵只能以90的倍数旋转，所以使参数对应于其他可能的东西更为直观由所使用的数据结构准确表示。 int是完美的，因为它可以告诉你将它旋转到某个单位（90）以及方向的次数。 double也可以告诉你，但它也包括此操作不支持的值（这本身就是违反直觉的）。
答案 42 :(得分：0)
所有当前的解决方案都有O（n ^ 2）开销作为临时空间（这不包括那些污秽的OOP骗子！）。这是一个使用O（1）内存使用的解决方案，将矩阵向右旋转90度。螺杆伸缩性，这个吸盘快速运行！
#include <algorithm>
#include <cstddef>
// Rotates an NxN matrix of type T 90 degrees to the right.
template <typename T, size_t N>
void rotate_matrix(T (&matrix)[N][N])
{
for(size_t i = 0; i < N; ++i)
for(size_t j = 0; j <= (N-i); ++j)
std::swap(matrix[i][j], matrix[j][i]);
}
免责声明：我实际上没有测试过这个。让我们玩弄错误！
答案 43 :(得分：0)
矩阵转置的C代码＆amp;旋转（+/- 90，+ / - 180）
`
#include <stdlib.h>
#include <memory.h>
#include <assert.h>
/*
Matrix transpose & rotate (+/-90, +/-180)
Supports both 2D arrays and 1D pointers with logical rows/cols
Supports square and non-square matrices, has in-place and copy features
See tests for examples of usage
tested gcc -std=c90 -Wall -pedantic, MSVC17
*/
typedef int matrix_data_t; /* matrix data type */
void transpose(const matrix_data_t* src, matrix_data_t* dst, int rows, int cols);
void transpose_inplace(matrix_data_t* data, int n );
void rotate(int direction, const matrix_data_t* src, matrix_data_t* dst, int rows, int cols);
void rotate_inplace(int direction, matrix_data_t* data, int n);
void reverse_rows(matrix_data_t* data, int rows, int cols);
void reverse_cols(matrix_data_t* data, int rows, int cols);
/* test/compare fn */
int test_cmp(const matrix_data_t* lhs, const matrix_data_t* rhs, int rows, int cols );
/* TESTS/USAGE */
void transpose_test() {
matrix_data_t sq3x3[9] = { 0,1,2,3,4,5,6,7,8 };/* 3x3 square, odd length side */
matrix_data_t sq3x3_cpy[9];
matrix_data_t sq3x3_2D[3][3] = { { 0,1,2 },{ 3,4,5 },{ 6,7,8 } };/* 2D 3x3 square */
matrix_data_t sq3x3_2D_copy[3][3];
/* expected test values */
const matrix_data_t sq3x3_orig[9] = { 0,1,2,3,4,5,6,7,8 };
const matrix_data_t sq3x3_transposed[9] = { 0,3,6,1,4,7,2,5,8};
matrix_data_t sq4x4[16]= { 0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15 };/* 4x4 square, even length*/
const matrix_data_t sq4x4_orig[16] = { 0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15 };
const matrix_data_t sq4x4_transposed[16] = { 0,4,8,12,1,5,9,13,2,6,10,14,3,7,11,15 };
/* 2x3 rectangle */
const matrix_data_t r2x3_orig[6] = { 0,1,2,3,4,5 };
const matrix_data_t r2x3_transposed[6] = { 0,3,1,4,2,5 };
matrix_data_t r2x3_copy[6];
matrix_data_t r2x3_2D[2][3] = { {0,1,2},{3,4,5} }; /* 2x3 2D rectangle */
matrix_data_t r2x3_2D_t[3][2];
/* matrix_data_t r3x2[6] = { 0,1,2,3,4,5 }; */
matrix_data_t r3x2_copy[6];
/* 3x2 rectangle */
const matrix_data_t r3x2_orig[6] = { 0,1,2,3,4,5 };
const matrix_data_t r3x2_transposed[6] = { 0,2,4,1,3,5 };
matrix_data_t r6x1[6] = { 0,1,2,3,4,5 }; /* 6x1 */
matrix_data_t r6x1_copy[6];
matrix_data_t r1x1[1] = { 0 }; /*1x1*/
matrix_data_t r1x1_copy[1];
/* 3x3 tests, 2D array tests */
transpose_inplace(sq3x3, 3); /* transpose in place */
assert(!test_cmp(sq3x3, sq3x3_transposed, 3, 3));
transpose_inplace(sq3x3, 3); /* transpose again */
assert(!test_cmp(sq3x3, sq3x3_orig, 3, 3));
transpose(sq3x3, sq3x3_cpy, 3, 3); /* transpose copy 3x3*/
assert(!test_cmp(sq3x3_cpy, sq3x3_transposed, 3, 3));
transpose((matrix_data_t*)sq3x3_2D, (matrix_data_t*)sq3x3_2D_copy, 3, 3); /* 2D array transpose/copy */
assert(!test_cmp((matrix_data_t*)sq3x3_2D_copy, sq3x3_transposed, 3, 3));
transpose_inplace((matrix_data_t*)sq3x3_2D_copy, 3); /* 2D array transpose in place */
assert(!test_cmp((matrix_data_t*)sq3x3_2D_copy, sq3x3_orig, 3, 3));
/* 4x4 tests */
transpose_inplace(sq4x4, 4); /* transpose in place */
assert(!test_cmp(sq4x4, sq4x4_transposed, 4,4));
transpose_inplace(sq4x4, 4); /* transpose again */
assert(!test_cmp(sq4x4, sq4x4_orig, 3, 3));
/* 2x3,3x2 tests */
transpose(r2x3_orig, r2x3_copy, 2, 3);
assert(!test_cmp(r2x3_copy, r2x3_transposed, 3, 2));
transpose(r3x2_orig, r3x2_copy, 3, 2);
assert(!test_cmp(r3x2_copy, r3x2_transposed, 2,3));
/* 2D array */
transpose((matrix_data_t*)r2x3_2D, (matrix_data_t*)r2x3_2D_t, 2, 3);
assert(!test_cmp((matrix_data_t*)r2x3_2D_t, r2x3_transposed, 3,2));
/* Nx1 test, 1x1 test */
transpose(r6x1, r6x1_copy, 6, 1);
assert(!test_cmp(r6x1_copy, r6x1, 1, 6));
transpose(r1x1, r1x1_copy, 1, 1);
assert(!test_cmp(r1x1_copy, r1x1, 1, 1));
}
void rotate_test() {
/* 3x3 square */
const matrix_data_t sq3x3[9] = { 0,1,2,3,4,5,6,7,8 };
const matrix_data_t sq3x3_r90[9] = { 6,3,0,7,4,1,8,5,2 };
const matrix_data_t sq3x3_180[9] = { 8,7,6,5,4,3,2,1,0 };
const matrix_data_t sq3x3_l90[9] = { 2,5,8,1,4,7,0,3,6 };
matrix_data_t sq3x3_copy[9];
/* 3x3 square, 2D */
matrix_data_t sq3x3_2D[3][3] = { { 0,1,2 },{ 3,4,5 },{ 6,7,8 } };
/* 4x4, 2D */
matrix_data_t sq4x4[4][4] = { { 0,1,2,3 },{ 4,5,6,7 },{ 8,9,10,11 },{ 12,13,14,15 } };
matrix_data_t sq4x4_copy[4][4];
const matrix_data_t sq4x4_r90[16] = { 12,8,4,0,13,9,5,1,14,10,6,2,15,11,7,3 };
const matrix_data_t sq4x4_l90[16] = { 3,7,11,15,2,6,10,14,1,5,9,13,0,4,8,12 };
const matrix_data_t sq4x4_180[16] = { 15,14,13,12,11,10,9,8,7,6,5,4,3,2,1,0 };
matrix_data_t r6[6] = { 0,1,2,3,4,5 }; /* rectangle with area of 6 (1x6,2x3,3x2, or 6x1) */
matrix_data_t r6_copy[6];
const matrix_data_t r1x6_r90[6] = { 0,1,2,3,4,5 };
const matrix_data_t r1x6_l90[6] = { 5,4,3,2,1,0 };
const matrix_data_t r1x6_180[6] = { 5,4,3,2,1,0 };
const matrix_data_t r2x3_r90[6] = { 3,0,4,1,5,2 };
const matrix_data_t r2x3_l90[6] = { 2,5,1,4,0,3 };
const matrix_data_t r2x3_180[6] = { 5,4,3,2,1,0 };
const matrix_data_t r3x2_r90[6] = { 4,2,0,5,3,1 };
const matrix_data_t r3x2_l90[6] = { 1,3,5,0,2,4 };
const matrix_data_t r3x2_180[6] = { 5,4,3,2,1,0 };
const matrix_data_t r6x1_r90[6] = { 5,4,3,2,1,0 };
const matrix_data_t r6x1_l90[6] = { 0,1,2,3,4,5 };
const matrix_data_t r6x1_180[6] = { 5,4,3,2,1,0 };
/* sq3x3 tests */
rotate(90, sq3x3, sq3x3_copy, 3, 3); /* +90 */
assert(!test_cmp(sq3x3_copy, sq3x3_r90, 3, 3));
rotate(-90, sq3x3, sq3x3_copy, 3, 3); /* -90 */
assert(!test_cmp(sq3x3_copy, sq3x3_l90, 3, 3));
rotate(180, sq3x3, sq3x3_copy, 3, 3); /* 180 */
assert(!test_cmp(sq3x3_copy, sq3x3_180, 3, 3));
/* sq3x3 in-place rotations */
memcpy( sq3x3_copy, sq3x3, 3 * 3 * sizeof(matrix_data_t));
rotate_inplace(90, sq3x3_copy, 3);
assert(!test_cmp(sq3x3_copy, sq3x3_r90, 3, 3));
rotate_inplace(-90, sq3x3_copy, 3);
assert(!test_cmp(sq3x3_copy, sq3x3, 3, 3)); /* back to 0 orientation */
rotate_inplace(180, sq3x3_copy, 3);
assert(!test_cmp(sq3x3_copy, sq3x3_180, 3, 3));
rotate_inplace(-180, sq3x3_copy, 3);
assert(!test_cmp(sq3x3_copy, sq3x3, 3, 3));
rotate_inplace(180, (matrix_data_t*)sq3x3_2D, 3);/* 2D test */
assert(!test_cmp((matrix_data_t*)sq3x3_2D, sq3x3_180, 3, 3));
/* sq4x4 */
rotate(90, (matrix_data_t*)sq4x4, (matrix_data_t*)sq4x4_copy, 4, 4);
assert(!test_cmp((matrix_data_t*)sq4x4_copy, sq4x4_r90, 4, 4));
rotate(-90, (matrix_data_t*)sq4x4, (matrix_data_t*)sq4x4_copy, 4, 4);
assert(!test_cmp((matrix_data_t*)sq4x4_copy, sq4x4_l90, 4, 4));
rotate(180, (matrix_data_t*)sq4x4, (matrix_data_t*)sq4x4_copy, 4, 4);
assert(!test_cmp((matrix_data_t*)sq4x4_copy, sq4x4_180, 4, 4));
/* r6 as 1x6 */
rotate(90, r6, r6_copy, 1, 6);
assert(!test_cmp(r6_copy, r1x6_r90, 1, 6));
rotate(-90, r6, r6_copy, 1, 6);
assert(!test_cmp(r6_copy, r1x6_l90, 1, 6));
rotate(180, r6, r6_copy, 1, 6);
assert(!test_cmp(r6_copy, r1x6_180, 1, 6));
/* r6 as 2x3 */
rotate(90, r6, r6_copy, 2, 3);
assert(!test_cmp(r6_copy, r2x3_r90, 2, 3));
rotate(-90, r6, r6_copy, 2, 3);
assert(!test_cmp(r6_copy, r2x3_l90, 2, 3));
rotate(180, r6, r6_copy, 2, 3);
assert(!test_cmp(r6_copy, r2x3_180, 2, 3));
/* r6 as 3x2 */
rotate(90, r6, r6_copy, 3, 2);
assert(!test_cmp(r6_copy, r3x2_r90, 3, 2));
rotate(-90, r6, r6_copy, 3, 2);
assert(!test_cmp(r6_copy, r3x2_l90, 3, 2));
rotate(180, r6, r6_copy, 3, 2);
assert(!test_cmp(r6_copy, r3x2_180, 3, 2));
/* r6 as 6x1 */
rotate(90, r6, r6_copy, 6, 1);
assert(!test_cmp(r6_copy, r6x1_r90, 6, 1));
rotate(-90, r6, r6_copy, 6, 1);
assert(!test_cmp(r6_copy, r6x1_l90, 6, 1));
rotate(180, r6, r6_copy, 6, 1);
assert(!test_cmp(r6_copy, r6x1_180, 6, 1));
}
/* test comparison fn, return 0 on match else non zero */
int test_cmp(const matrix_data_t* lhs, const matrix_data_t* rhs, int rows, int cols) {
int r, c;
for (r = 0; r < rows; ++r) {
for (c = 0; c < cols; ++c) {
if ((lhs + r * cols)[c] != (rhs + r * cols)[c])
return -1;
}
}
return 0;
}
/*
Reverse values in place of each row in 2D matrix data[rows][cols] or in 1D pointer with logical rows/cols
[A B C] -> [C B A]
[D E F] [F E D]
*/
void reverse_rows(matrix_data_t* data, int rows, int cols) {
int r, c;
matrix_data_t temp;
matrix_data_t* pRow = NULL;
for (r = 0; r < rows; ++r) {
pRow = (data + r * cols);
for (c = 0; c < (int)(cols / 2); ++c) { /* explicit truncate */
temp = pRow[c];
pRow[c] = pRow[cols - 1 - c];
pRow[cols - 1 - c] = temp;
}
}
}
/*
Reverse values in place of each column in 2D matrix data[rows][cols] or in 1D pointer with logical rows/cols
[A B C] -> [D E F]
[D E F] [A B C]
*/
void reverse_cols(matrix_data_t* data, int rows, int cols) {
int r, c;
matrix_data_t temp;
matrix_data_t* pRowA = NULL;
matrix_data_t* pRowB = NULL;
for (c = 0; c < cols; ++c) {
for (r = 0; r < (int)(rows / 2); ++r) { /* explicit truncate */
pRowA = data + r * cols;
pRowB = data + cols * (rows - 1 - r);
temp = pRowA[c];
pRowA[c] = pRowB[c];
pRowB[c] = temp;
}
}
}
/* Transpose NxM matrix to MxN matrix in O(n) time */
void transpose(const matrix_data_t* src, matrix_data_t* dst, int N, int M) {
int i;
for (i = 0; i<N*M; ++i) dst[(i%M)*N + (i / M)] = src[i]; /* one-liner version */
/*
expanded version of one-liner: calculate XY based on array index, then convert that to YX array index
int i,j,x,y;
for (i = 0; i < N*M; ++i) {
x = i % M;
y = (int)(i / M);
j = x * N + y;
dst[j] = src[i];
}
*/
/*
nested for loop version
using ptr arithmetic to get proper row/column
this is really just dst[col][row]=src[row][col]
int r, c;
for (r = 0; r < rows; ++r) {
for (c = 0; c < cols; ++c) {
(dst + c * rows)[r] = (src + r * cols)[c];
}
}
*/
}
/*
Transpose NxN matrix in place
*/
void transpose_inplace(matrix_data_t* data, int N ) {
int r, c;
matrix_data_t temp;
for (r = 0; r < N; ++r) {
for (c = r; c < N; ++c) { /*start at column=row*/
/* using ptr arithmetic to get proper row/column */
/* this is really just
temp=dst[col][row];
dst[col][row]=src[row][col];
src[row][col]=temp;
*/
temp = (data + c * N)[r];
(data + c * N)[r] = (data + r * N)[c];
(data + r * N)[c] = temp;
}
}
}
/*
Rotate 1D or 2D src matrix to dst matrix in a direction (90,180,-90)
Precondition: src and dst are 2d matrices with dimensions src[rows][cols] and dst[cols][rows] or 1D pointers with logical rows/cols
*/
void rotate(int direction, const matrix_data_t* src, matrix_data_t* dst, int rows, int cols) {
switch (direction) {
case -90:
transpose(src, dst, rows, cols);
reverse_cols(dst, cols, rows);
break;
case 90:
transpose(src, dst, rows, cols);
reverse_rows(dst, cols, rows);
break;
case 180:
case -180:
/* bit copy to dst, use in-place reversals */
memcpy(dst, src, rows*cols*sizeof(matrix_data_t));
reverse_cols(dst, cols, rows);
reverse_rows(dst, cols, rows);
break;
}
}
/*
Rotate array in a direction.
Array must be NxN 2D or 1D array with logical rows/cols
Direction can be (90,180,-90,-180)
*/
void rotate_inplace( int direction, matrix_data_t* data, int n) {
switch (direction) {
case -90:
transpose_inplace(data, n);
reverse_cols(data, n, n);
break;
case 90:
transpose_inplace(data, n);
reverse_rows(data, n, n);
break;
case 180:
case -180:
reverse_cols(data, n, n);
reverse_rows(data, n, n);
break;
}
}
`
答案 44 :(得分：0)
#include <iostream>
#include <iomanip>
using namespace std;
const int SIZE=3;
void print(int a[][SIZE],int);
void rotate(int a[][SIZE],int);
void main()
{
int a[SIZE][SIZE]={{11,22,33},{44,55,66},{77,88,99}};
cout<<"the array befor rotate\n";
print(a,SIZE);
rotate( a,SIZE);
cout<<"the array after rotate\n";
print(a,SIZE);
cout<<endl;
}
void print(int a[][SIZE],int SIZE)
{
int i,j;
for(i=0;i<SIZE;i++)
for(j=0;j<SIZE;j++)
cout<<a[i][j]<<setw(4);
}
void rotate(int a[][SIZE],int SIZE)
{
int temp[3][3],i,j;
for(i=0;i<SIZE;i++)
for(j=0;j<SIZE/2.5;j++)
{
temp[i][j]= a[i][j];
a[i][j]= a[j][SIZE-i-1] ;
a[j][SIZE-i-1] =temp[i][j];
}
}
答案 45 :(得分：0)
这是一个Javascript解决方案：
const transpose = m => m[0].map((x,i) => m.map(x => x[i]));
a: // original matrix
123
456
789
transpose(a).reverse(); // rotate 90 degrees counter clockwise
369
258
147
transpose(a.slice().reverse()); // rotate 90 degrees clockwise
741
852
963
transpose(transpose(a.slice().reverse()).slice().reverse())
// rotate 180 degrees
987
654
321
答案 46 :(得分：-1)
public static void rotateMatrix(int[,] matrix)
{
//C#, to rotate an N*N matrix in place
int n = matrix.GetLength(0);
int layers = n / 2;
int temp, temp2;
for (int i = 0; i < layers; i++) // for a 5 * 5 matrix, layers will be 2, since at layer three there would be only one element, (2,2), and we do not need to rotate it with itself
{
int offset = 0;
while (offset < n - 2 * i - 1)
{
// top right <- top left
temp = matrix[i + offset, n - i - 1]; //top right value when offset is zero
matrix[i + offset, n - i - 1] = matrix[i, i + offset];
//bottom right <- top right
temp2 = matrix[n - i - 1, n - i - 1 - offset]; //bottom right value when offset is zero
matrix[n - i - 1, n - i - 1 - offset] = temp;
//bottom left <- bottom right
temp = matrix[n - i - 1 - offset, i];
matrix[n - i - 1 - offset, i] = temp2;
//top left <- bottom left
matrix[i, i + offset] = temp;
offset++;
}
}
}
答案 47 :(得分：-1)
在Java中
public class Matrix {
/* Author Shrikant Dande */
private static void showMatrix(int[][] arr,int rows,int col){
for(int i =0 ;i<rows;i++){
for(int j =0 ;j<col;j++){
System.out.print(arr[i][j]+" ");
}
System.out.println();
}
}
private static void rotateMatrix(int[][] arr,int rows,int col){
int[][] tempArr = new int[4][4];
for(int i =0 ;i<rows;i++){
for(int j =0 ;j<col;j++){
tempArr[i][j] = arr[rows-1-j][i];
System.out.print(tempArr[i][j]+" ");
}
System.out.println();
}
}
public static void main(String[] args) {
int[][] arr = { {1, 2, 3, 4},
{5, 6, 7, 8},
{9, 1, 2, 5},
{7, 4, 8, 9}};
int rows = 4,col = 4;
showMatrix(arr, rows, col);
System.out.println("------------------------------------------------");
rotateMatrix(arr, rows, col);
}
}
答案 48 :(得分：-1)
可以非常干净地递归完成，这是我在golang中的实现！
在go golang中以递归方式旋转nxn矩阵而没有额外的内存
func rot90(a [][]int) {
n := len(a)
if n == 1 {
return
}
for i := 0; i < n; i++ {
a[0][i], a[n-1-i][n-1] = a[n-1-i][n-1], a[0][i]
}
rot90(a[1:])
}
答案 49 :(得分：-1)
对于原位旋转，不可能比O（n ^ 2）更快，因为如果我们想要旋转矩阵，我们必须至少触摸所有n ^ 2元素一次，否无论你正在实施什么算法。
答案 50 :(得分：-1)
#!/usr/bin/env python
original = [ [1,2,3],
[4,5,6],
[7,8,9] ]
# Rotate matrix 90 degrees...
for i in map(None,*original[::-1]):
print str(i) + '\n'
这会使侧面旋转90度（即123（顶侧）现在是741（左侧）。
这个Python解决方案有效，因为它使用切片和负步骤来反转行顺序（将7引入顶部）
original = [ [7,8,9],
[4,5,6],
[1,2,3] ]
然后使用map（以及隐含的身份函数，它是作为第一个arg的None的映射结果）以及*来按顺序解包所有元素，以重新组合列（即，第一个元素被放入一个元组在一起，第二个元素被放在一个元组中，等等）。您有效地获得了以下重新组合：
original = [[7,8,9],
[4,5,6],
[1,2,3]]
答案 51 :(得分：-1)
这是一个C＃静态泛型方法，可以为您完成工作。变量有名，所以你可以很容易地理解algorythm。
private static T[,] Rotate180 <T> (T[,] matrix)
{
var height = matrix.GetLength (0);
var width = matrix.GetLength (1);
var answer = new T[height, width];
for (int y = 0; y < height / 2; y++)
{
int topY = y;
int bottomY = height - 1 - y;
for (int topX = 0; topX < width; topX++)
{
var bottomX = width - topX - 1;
answer[topY, topX] = matrix[bottomY, bottomX];
answer[bottomY, bottomX] = matrix[topY, topX];
}
}
if (height % 2 == 0)
return answer;
var centerY = height / 2;
for (int leftX = 0; leftX < Mathf.CeilToInt(width / 2f); leftX++)
{
var rightX = width - 1 - leftX;
answer[centerY, leftX] = matrix[centerY, rightX];
answer[centerY, rightX] = matrix[centerY, leftX];
}
return answer;
}
答案 52 :(得分：-1)
我的 C＃示例代码，用于@dimple发送的优秀算法：
/* 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
*/
答案 53 :(得分：-1)
这是将数组旋转90度的简单C代码。希望这可以帮助。
#include <stdio.h>
void main(){
int arr[3][4] = {85, 2, 85, 4,
85, 6, 7, 85,
9, 85, 11, 12};
int arr1[4][3];
int i = 0, j = 0;
for(i=0;i<4;i++){
int k = 2;//k = (number of columns in the new array arr1 - 1)
for(j=0;j<3;j++){
arr1[i][j]=arr[k][i];
k--;
}
}
int l, m;
for(l=0;l<4;l++){
for(m=0;m<3;m++){
printf("%d ", arr1[l][m]);
}
printf("\n");
}
}//end main
答案 54 :(得分：-1)
这是Java：
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;
}
}
}
答案 55 :(得分：-1)
这是一个被高估的面试问题。
我的建议是：不要让面试官对你解决这个问题的疯狂建议感到困惑。使用白板绘制输入数组的索引，然后绘制输出数组的索引。旋转前后的列索引样本如下所示：
30 --> 00
20 --> 01
10 --> 02
00 --> 03
31 --> 10
21 --> 11
11 --> 12
01 --> 13
注意旋转后的数字模式。
下面提供了一个简洁的Java解决方案。它经过测试，有效：
Input:
M A C P
B N L D
Y E T S
I W R Z
Output:
I Y B M
W E N A
R T L C
Z S D P
/**
* (c) @author "G A N MOHIM"
* Oct 3, 2015
* RotateArrayNintyDegree.java
*/
package rotatearray;
public class RotateArrayNintyDegree {
public char[][] rotateArrayNinetyDegree(char[][] input) {
int k; // k is used to generate index for output array
char[][] output = new char[input.length] [input[0].length];
for (int i = 0; i < input.length; i++) {
k = 0;
for (int j = input.length-1; j >= 0; j--) {
output[i][k] = input[j][i]; // note how i is used as column index, and j as row
k++;
}
}
return output;
}
public void printArray(char[][] charArray) {
for (int i = 0; i < charArray.length; i++) {
for (int j = 0; j < charArray[0].length; j++) {
System.out.print(charArray[i][j] + " ");
}
System.out.println();
}
}
public static void main(String[] args) {
char[][] input =
{ {'M', 'A', 'C', 'P'},
{'B', 'N', 'L', 'D'},
{'Y', 'E', 'T', 'S'},
{'I', 'W', 'R', 'Z'}
};
char[][] output = new char[input.length] [input[0].length];
RotateArrayNintyDegree rotationObj = new RotateArrayNintyDegree();
rotationObj.printArray(input);
System.out.println("\n");
output = rotationObj.rotateArrayNinetyDegree(input);
rotationObj.printArray(output);
}
}
答案 56 :(得分：-1)
使用矢量矢量进行顺时针90度旋转。
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
//Rotate a Matrix by 90 degrees
void rotateMatrix(vector<vector<int> > &matrix){
int n=matrix.size();
for(int i=0;i<n;i++){
for(int j=i+1;j<n;j++){
swap(matrix[i][j],matrix[j][i]);
}
}
for(int i=0;i<n;i++){
reverse(matrix[i].begin(),matrix[i].end());
}
}
int main(){
int n;
cout<<"enter the size of the matrix:"<<endl;
while (cin >> n) {
vector< vector<int> > m;
cout<<"enter the elements"<<endl;
for (int i = 0; i < n; i++) {
m.push_back(vector<int>(n));
for (int j = 0; j < n; j++)
scanf("%d", &m[i][j]);
}
cout<<"the rotated matrix is:"<<endl;
rotateMatrix(m);
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++)
cout << m[i][j] << ' ';
cout << endl;
}
}
return 0;
}
答案 57 :(得分：-1)
/* 90-degree clockwise:
temp_array = left_col
left_col = bottom_row
bottom_row = reverse(right_col)
reverse(right_col) = reverse(top_row)
reverse(top_row) = temp_array
*/
void RotateClockwise90(int ** arr, int lo, int hi) {
if (lo >= hi)
return;
for (int i=lo; i<hi; i++) {
int j = lo+hi-i;
int temp = arr[i][lo];
arr[i][lo] = arr[hi][i];
arr[hi][i] = arr[j][hi];
arr[j][hi] = arr[lo][j];
arr[lo][j] = temp;
}
RotateClockwise90(arr, lo+1, hi-1);
}
答案 58 :(得分：-1)
将矩阵旋转90度的JavaScript解决方案：
function rotateBy90(m) {
var length = m.length;
//for each layer of the matrix
for (var first = 0; first < length >> 1; first++) {
var last = length - 1 - first;
for (var i = first; i < last; i++) {
var top = m[first][i]; //store top
m[first][i] = m[last - i][first]; //top = left
m[last - i][first] = m[last][last - i]; //left = bottom
m[last][last - i] = m[i][last]; //bottom = right
m[i][last] = top; //right = top
}
}
return m;
}
答案 59 :(得分：-1)
对于新手程序员，使用普通的C ++。 （Borland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;
}
答案 60 :(得分：-1)
<强> PHP：
array_unshift($array, null);
$array = call_user_func_array("array_map", $array);
如果您需要在90度旋转矩形二维数组，请在上面的代码之前或之后添加以下行（取决于您需要的旋转方向）：
$array = array_reverse($array);
答案 61 :(得分：-2)
试试我的图书馆AbacusUtil：
@Test
public void test_42519() throws Exception {
final IntMatrix matrix = IntMatrix.range(0, 16).reshape(4);
N.println("======= original =======================");
matrix.println();
// print out:
// [0, 1, 2, 3]
// [4, 5, 6, 7]
// [8, 9, 10, 11]
// [12, 13, 14, 15]
N.println("======= rotate 90 ======================");
matrix.rotate90().println();
// print out:
// [12, 8, 4, 0]
// [13, 9, 5, 1]
// [14, 10, 6, 2]
// [15, 11, 7, 3]
N.println("======= rotate 180 =====================");
matrix.rotate180().println();
// print out:
// [15, 14, 13, 12]
// [11, 10, 9, 8]
// [7, 6, 5, 4]
// [3, 2, 1, 0]
N.println("======= rotate 270 ======================");
matrix.rotate270().println();
// print out:
// [3, 7, 11, 15]
// [2, 6, 10, 14]
// [1, 5, 9, 13]
// [0, 4, 8, 12]
N.println("======= transpose =======================");
matrix.transpose().println();
// print out:
// [0, 4, 8, 12]
// [1, 5, 9, 13]
// [2, 6, 10, 14]
// [3, 7, 11, 15]
final IntMatrix bigMatrix = IntMatrix.range(0, 10000_0000).reshape(10000);
// It take about 2 seconds to rotate 10000 X 10000 matrix.
Profiler.run(1, 2, 3, "sequential", () -> bigMatrix.rotate90()).printResult();
// Want faster? Go parallel. 1 second to rotate 10000 X 10000 matrix.
final int[][] a = bigMatrix.array();
final int[][] c = new int[a[0].length][a.length];
final int n = a.length;
final int threadNum = 4;
Profiler.run(1, 2, 3, "parallel", () -> {
IntStream.range(0, n).parallel(threadNum).forEach(i -> {
for (int j = 0; j < n; j++) {
c[i][j] = a[n - j - 1][i];
}
});
}).printResult();
}
答案 62 :(得分：-2)
O（1）记忆算法：
旋转最外面的数据，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]
要进行此轮换，我们知道
dest[j][n-1-i] = src[i][j]
请注意以下内容： a（0,0） - ＆gt;一个（0,3） a（0,3） - ＆gt;一个（3,3） a（3,3） - ＆gt;一个（3,0） a（3,0） - ＆gt;一个（0,0）
因此它是一个圆圈，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面的元素。
因此我们可以如下结束：
function rotate(array, N)
{
Rotate outer-most data
rotate a new array with N-2 or you can do the similar action following step1
}