将矩阵顺时针方向旋转 90 度而不使用任何额外空间

将矩阵顺时针旋转 90 度而不使用任何额外空间

给定一个方阵，在不使用任何额外空间的情况下，将其顺时针方向旋转 90 度。

例子：

Input:
1 2 3 
4 5 6
7 8 9  
Output:
7 4 1 
8 5 2
9 6 3

Input:
1 2
3 4
Output:
3 1
4 2 

方法一

方法：方法类似于 Inplace 旋转方阵 90 度 |设置 1。唯一不同的是以顺时针方向打印循环的元素，即 N x N 矩阵将具有 floor(N/2) 平方循环。
例如，一个 3 X 3 矩阵将有 1 个循环。循环由其第一行、最后一列、最后一行和第一列构成。
对于每个方形循环，我们以顺时针方向交换矩阵中与相应单元格相关的元素。为此，我们只需要一个临时变量。

解释：

Let size of row and column be 3. 
During first iteration – 
a[i][j] = Element at first index (leftmost corner top)= 1.
a[j][n-1-i]= Rightmost corner top Element = 3.
a[n-1-i][n-1-j] = Rightmost corner bottom element = 9.
a[n-1-j][i] = Leftmost corner bottom element = 7.
Move these elements in the clockwise direction. 
During second iteration – 
a[i][j] = 2.
a[j][n-1-j] = 6.
a[n-1-i][n-1-j] = 8.
a[n-1-j][i] = 4. 
Similarly, move these elements in the clockwise direction. 

下面是上述方法的实现：

C++

// C++ implementation of above approach
#include 
using namespace std;
 
#define N 4
 
// Function to rotate the matrix 90 degree clockwise
void rotate90Clockwise(int a[N][N])
{
 
    // Traverse each cycle
    for (int i = 0; i < N / 2; i++) {
        for (int j = i; j < N - i - 1; j++) {
 
            // Swap elements of each cycle
            // in clockwise direction
            int temp = a[i][j];
            a[i][j] = a[N - 1 - j][i];
            a[N - 1 - j][i] = a[N - 1 - i][N - 1 - j];
            a[N - 1 - i][N - 1 - j] = a[j][N - 1 - i];
            a[j][N - 1 - i] = temp;
        }
    }
}
 
// Function for print matrix
void printMatrix(int arr[N][N])
{
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < N; j++)
            cout << arr[i][j] << " ";
        cout << '\n';
    }
}
 
// Driver code
int main()
{
    int arr[N][N] = { { 1, 2, 3, 4 },
                      { 5, 6, 7, 8 },
                      { 9, 10, 11, 12 },
                      { 13, 14, 15, 16 } };
    rotate90Clockwise(arr);
    printMatrix(arr);
    return 0;
}

Java

// Java implementation of above approach
import java.io.*;
 
class GFG
{
     
static int N = 4;
 
// Function to rotate the matrix 90 degree clockwise
static void rotate90Clockwise(int a[][])
{
 
    // Traverse each cycle
    for (int i = 0; i < N / 2; i++)
    {
        for (int j = i; j < N - i - 1; j++)
        {
 
            // Swap elements of each cycle
            // in clockwise direction
            int temp = a[i][j];
            a[i][j] = a[N - 1 - j][i];
            a[N - 1 - j][i] = a[N - 1 - i][N - 1 - j];
            a[N - 1 - i][N - 1 - j] = a[j][N - 1 - i];
            a[j][N - 1 - i] = temp;
        }
    }
}
 
// Function for print matrix
static void printMatrix(int arr[][])
{
    for (int i = 0; i < N; i++)
    {
        for (int j = 0; j < N; j++)
        System.out.print( arr[i][j] + " ");
        System.out.println();
    }
}
 
// Driver code
 
    public static void main (String[] args)
    {
            int arr[][] = { { 1, 2, 3, 4 },
                      { 5, 6, 7, 8 },
                      { 9, 10, 11, 12 },
                      { 13, 14, 15, 16 } };
    rotate90Clockwise(arr);
    printMatrix(arr);
    }
}
 
// This code has been contributed by inder_verma.

Python

# Function to rotate the matrix
# 90 degree clockwise
def rotate90Clockwise(A):
    N = len(A[0])
    for i in range(N // 2):
        for j in range(i, N - i - 1):
            temp = A[i][j]
            A[i][j] = A[N - 1 - j][i]
            A[N - 1 - j][i] = A[N - 1 - i][N - 1 - j]
            A[N - 1 - i][N - 1 - j] = A[j][N - 1 - i]
            A[j][N - 1 - i] = temp
 
# Function to print the matrix
def printMatrix(A):
    N = len(A[0])
    for i in range(N):
        print(A[i])
 
# Driver code
A = [[1, 2, 3, 4],
     [5, 6, 7, 8],
     [9, 10, 11, 12],
     [13, 14, 15, 16]]
rotate90Clockwise(A)
printMatrix(A)
 
# This code was contributed
# by pk_tautolo

C#

// C# implementation of above approach
using System;
 
class GFG
{
static int N = 4;
 
// Function to rotate the matrix
// 90 degree clockwise
static void rotate90Clockwise(int[,] a)
{
 
    // Traverse each cycle
    for (int i = 0; i < N / 2; i++)
    {
        for (int j = i; j < N - i - 1; j++)
        {
 
            // Swap elements of each cycle
            // in clockwise direction
            int temp = a[i, j];
            a[i, j] = a[N - 1 - j, i];
            a[N - 1 - j, i] = a[N - 1 - i, N - 1 - j];
            a[N - 1 - i, N - 1 - j] = a[j, N - 1 - i];
            a[j, N - 1 - i] = temp;
        }
    }
}
 
// Function for print matrix
static void printMatrix(int[,] arr)
{
    for (int i = 0; i < N; i++)
    {
        for (int j = 0; j < N; j++)
        Console.Write( arr[i, j] + " ");
        Console.Write("\n");
    }
}
 
// Driver code
public static void Main ()
    {
    int [,]arr = {{1, 2, 3, 4},
                  {5, 6, 7, 8},
                  {9, 10, 11, 12},
                  {13, 14, 15, 16}};
    rotate90Clockwise(arr);
    printMatrix(arr);
}
}
 
// This code is contributed
// by ChitraNayal

Javascript

 
// Javascript implementation of above approach
 
    var N = 4;
 
    // Function to rotate the matrix 90 degree clockwise
    function rotate90Clockwise(a) {
 
        // Traverse each cycle
        for (i = 0; i < parseInt(N / 2); i++) {
            for (j = i; j < N - i - 1; j++) {
 
                // Swap elements of each cycle
                // in clockwise direction
                var temp = a[i][j];
                a[i][j] = a[N - 1 - j][i];
                a[N - 1 - j][i] = a[N - 1 - i][N - 1 - j];
                a[N - 1 - i][N - 1 - j] = a[j][N - 1 - i];
                a[j][N - 1 - i] = temp;
            }
        }
    }
 
    // Function for prvar matrix
    function printMatrix(arr) {
        for (i = 0; i < N; i++) {
            for (j = 0; j < N; j++)
                document.write(arr[i][j] + " ");
            document.write("<br/>");
        }
    }
 
    // Driver code
 
     
        var arr = [ [ 1, 2, 3, 4 ],
                    [ 5, 6, 7, 8 ],
                    [ 9, 10, 11, 12 ],
                    [ 13, 14, 15, 16 ] ];
        rotate90Clockwise(arr);
        printMatrix(arr);
 
// This code contributed by Rajput-Ji
 

---

---

输出

13 9 5 1 
14 10 6 2 
15 11 7 3 
16 12 8 4 

方法二：

方法：该方法基于旋转矩阵后由索引构成的模式。考虑下图可以清楚地了解它。

考虑具有如下索引 (i, j) 的 3 x 3 矩阵。

00 01 02
10 11 12
20 21 22

将矩阵顺时针旋转 90 度后，索引转换为
20 10 00 current_row_index = 0, i = 2, 1, 0
21 11 01 current_row_index = 1, i = 2, 1, 0
22 12 02 current_row_index = 2, i = 2, 1, 0

观察：在任何行中，对于每一个递减的行索引i ，都存在一个常量列索引j ， 使得j = current_row_index 。

可以使用 2 个嵌套循环打印此图案。
（这种写入索引的模式是通过写入所需元素的确切索引来实现的
它们实际存在于原始数组中的位置。）

下面是上述方法的实现：

C++

// C++ implementation of above approach
#include 
using namespace std;
 
#define N 4
 
// Function to rotate the matrix 90 degree clockwise
void rotate90Clockwise(int arr[N][N])
{
    // printing the matrix on the basis of
    // observations made on indices.
    for (int j = 0; j < N; j++)
    {
        for (int i = N - 1; i >= 0; i--)
            cout << arr[i][j] << " ";
        cout << '\n';
    }
}
 
// Driver code
int main()
{
    int arr[N][N] = { { 1, 2, 3, 4 },
                      { 5, 6, 7, 8 },
                      { 9, 10, 11, 12 },
                      { 13, 14, 15, 16 } };
    rotate90Clockwise(arr);
    return 0;
}
 
// This code is contributed by yashbeersingh42

Java

// Java implementation of above approach
import java.io.*;
 
class GFG {
    static int N = 4;
 
    // Function to rotate the matrix 90 degree clockwise
    static void rotate90Clockwise(int arr[][])
    {
        // printing the matrix on the basis of
        // observations made on indices.
        for (int j = 0; j < N; j++)
        {
            for (int i = N - 1; i >= 0; i--)
                System.out.print(arr[i][j] + " ");
            System.out.println();
        }
    }
    public static void main(String[] args)
    {
        int arr[][] = { { 1, 2, 3, 4 },
                        { 5, 6, 7, 8 },
                        { 9, 10, 11, 12 },
                        { 13, 14, 15, 16 } };
        rotate90Clockwise(arr);
    }
}
// This code is contributed by yashbeersingh42

蟒蛇3

# Python3 implementation of above approach
N = 4
 
# Function to rotate the matrix 90 degree clockwise
def rotate90Clockwise(arr) :
    global N
     
    # printing the matrix on the basis of
    # observations made on indices.
    for j in range(N) :
        for i in range(N - 1, -1, -1) :
            print(arr[i][j], end = " ")
        print()
         
# Driver code       
arr = [ [ 1, 2, 3, 4 ],
          [ 5, 6, 7, 8 ],
          [ 9, 10, 11, 12 ],
          [ 13, 14, 15, 16 ] ]
rotate90Clockwise(arr);
 
# This code is contributed by divyesh072019.

C＃

// C# implementation of above approach
using System;
class GFG {
     
    static int N = 4;
     
    // Function to rotate the matrix 90 degree clockwise
    static void rotate90Clockwise(int[,] arr)
    {
       
        // printing the matrix on the basis of
        // observations made on indices.
        for (int j = 0; j < N; j++)
        {
            for (int i = N - 1; i >= 0; i--)
                Console.Write(arr[i, j] + " ");
            Console.WriteLine();
        }
    }
     
  // Driver code
  static void Main() {
    int[,] arr = { { 1, 2, 3, 4 },
                  { 5, 6, 7, 8 },
                  { 9, 10, 11, 12 },
                  { 13, 14, 15, 16 } };
    rotate90Clockwise(arr);
  }
}
 
// This code is contributed by divyeshrabadiya07.

Javascript

// javascript implementation of above approach
    var N = 4;
 
    // Function to rotate the matrix 90 degree clockwise
    function rotate90Clockwise(arr) {
        // printing the matrix on the basis of
        // observations made on indices.
        for (j = 0; j < N; j++) {
            for (i = N - 1; i >= 0; i--)
                document.write(arr[i][j] + " ");
            document.write("<br/>");
        }
    }
 
     
        var arr = [ [ 1, 2, 3, 4 ],
                    [ 5, 6, 7, 8 ],
                    [ 9, 10, 11, 12 ],
                    [ 13, 14, 15, 16 ] ];
        rotate90Clockwise(arr);
 
// This code contributed by Rajput-Ji

输出

13 9 5 1 
14 10 6 2 
15 11 7 3 
16 12 8 4 

方法三：

方法：方法是将给定的矩阵旋转两次，第一次相对于主对角线，下一次相对于中间列旋转合成矩阵，考虑下图可以清楚地了解它。

将方阵顺时针旋转 90 度

下面是上述方法的实现：

C++

#include 
using namespace std;
 
#define N 4
 
void print(int arr[N][N])
{
    for(int i = 0; i < N; ++i)
    {
        for(int j = 0; j < N; ++j)
            cout << arr[i][j] << " ";
             
        cout << '\n';
    }
}
 
void rotate(int arr[N][N])
{
     
    // First rotation
    // with respect to main diagonal
    for(int i = 0; i < N; ++i)
    {
        for(int j = 0; j < i; ++j)
        {
            int temp = arr[i][j];
            arr[i][j] = arr[j][i];
            arr[j][i] = temp;
        }
    }
     
    // Second rotation
    // with respect to middle column
    for(int i = 0; i < N; ++i)
    {
        for(int j = 0; j < N / 2; ++j)
        {
            int temp = arr[i][j];
            arr[i][j] = arr[i][N - j - 1];
            arr[i][N - j - 1] = temp;
        }
    }
}
 
// Driver code
int main()
{
    int arr[N][N] = { { 1, 2, 3, 4 },
                      { 5, 6, 7, 8 },
                      { 9, 10, 11, 12 },
                      { 13, 14, 15, 16 } };
    rotate(arr);
    print(arr);
    return 0;
}
 
// This code is contributed  by Rahul Verma

Java

import java.io.*;
 
class GFG {
   
  static void rotate(int[][] arr) {
 
        int n=arr.length;
     
    // first rotation
    // with respect to main diagonal
        for(int i=0;i

C＃

using System;
using System.Collections.Generic;
public class GFG {
   
  static void rotate(int[,] arr) {
 
        int n=arr.GetLength(0);
     
    // first rotation
    // with respect to main diagonal
        for(int i=0;i

Javascript

 
let N = 4
 
function print(arr)
{
    for(let i = 0; i < N; ++i)
    {
        for(let j = 0; j < N; ++j)
            document.write(arr[i][j] + " ");
             
        document.write("<br>");
    }
}
 
function rotate(arr)
{
     
    // First rotation
    // with respect to main diagonal
    for(let i = 0; i < N; ++i)
    {
        for(let j = 0; j < i; ++j)
        {
            let temp = arr[i][j];
            arr[i][j] = arr[j][i];
            arr[j][i] = temp;
        }
    }
     
    // Second rotation
    // with respect to middle column
    for(let i = 0; i < N; ++i)
    {
        for(let j = 0; j < N / 2; ++j)
        {
            let temp = arr[i][j];
            arr[i][j] = arr[i][N - j - 1];
            arr[i][N - j - 1] = temp;
        }
    }
}
 
// Driver code
 
    let arr = [ [ 1, 2, 3, 4 ],
                    [ 5, 6, 7, 8 ],
                    [ 9, 10, 11, 12 ],
                    [ 13, 14, 15, 16 ] ];
    rotate(arr);
    print(arr);
     
//This code is contributed by Mayank Tyagi

方法四：

方法：这种方法与方法 3 类似，唯一的区别是在第一次旋转时，我们围绕第二对角线旋转，然后围绕中间行旋转。

将方阵顺时针旋转 90 度

下面是上述方法的实现：

C++

#include 
using namespace std;
 
#define N 4
 
void print(int arr[N][N])
{
    for(int i = 0; i < N; ++i)
    {
        for(int j = 0; j < N; ++j)
            cout << arr[i][j] << " ";
             
        cout << '\n';
    }
}
 
void rotate(int arr[N][N])
{
     
    // First rotation
    // with respect to Secondary diagonal
    for(int i = 0; i < N; i++)
    {
        for(int j = 0; j < N - i; j++)
        {
            int temp = arr[i][j];
            arr[i][j] = arr[N - 1 - j][N - 1 - i];
            arr[N - 1 - j][N - 1 - i] = temp;
        }
    }
     
    // Second rotation
    // with respect to middle row
    for(int i = 0; i < N / 2; i++)
    {
        for(int j = 0; j < N; j++)
        {
            int temp = arr[i][j];
            arr[i][j] = arr[N - 1 - i][j];
            arr[N - 1 - i][j] = temp;
        }
    }
}
 
// Driver code
int main()
{
    int arr[N][N] = { { 1, 2, 3, 4 },
                      { 5, 6, 7, 8 },
                      { 9, 10, 11, 12 },
                      { 13, 14, 15, 16 } };
    rotate(arr);
    print(arr);
    return 0;
}
 
// This code is contributed  by Rahul Verma

Java

import java.io.*;
 
class GFG {
 
    static void rotate(int[][] arr)
    {
 
        int n = arr.length;
 
        // first rotation
        // with respect to Secondary diagonal
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n - i; j++) {
                int temp = arr[i][j];
                arr[i][j] = arr[n - 1 - j][n - 1 - i];
                arr[n - 1 - j][n - 1 - i] = temp;
            }
        }
        // Second rotation
        // with respect to middle row
        for (int i = 0; i < n / 2; i++) {
            for (int j = 0; j < n; j++) {
                int temp = arr[i][j];
                arr[i][j] = arr[n - 1 - i][j];
                arr[n - 1 - i][j] = temp;
            }
        }
    }
 
    // to print matrix
    static void printMatrix(int arr[][])
    {
        int n = arr.length;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++)
                System.out.print(arr[i][j] + " ");
            System.out.println();
        }
    }
    // Driver code
    public static void main(String[] args)
    {
        int arr[][] = { { 1, 2, 3, 4 },
                        { 5, 6, 7, 8 },
                        { 9, 10, 11, 12 },
                        { 13, 14, 15, 16 } };
        rotate(arr);
        printMatrix(arr);
    }
}
// This code is contributed  by Rahul Verma

C＃

using System;
using System.Collections.Generic;
 
class GFG{
 
static void rotate(int[,] arr)
{
    int n = arr.GetLength(0);
 
    // First rotation
    // with respect to Secondary diagonal
    for(int i = 0; i < n; i++)
    {
        for(int j = 0; j < n - i; j++)
        {
            int temp = arr[i, j];
            arr[i, j] = arr[n - 1 - j, n - 1 - i];
            arr[n - 1 - j, n - 1 - i] = temp;
        }
    }
     
    // Second rotation
    // with respect to middle row
    for(int i = 0; i < n / 2; i++)
    {
        for(int j = 0; j < n; j++)
        {
            int temp = arr[i, j];
            arr[i, j] = arr[n - 1 - i, j];
            arr[n - 1 - i, j] = temp;
        }
    }
}
 
// To print matrix
static void printMatrix(int [,]arr)
{
    int n = arr.GetLength(0);
    for(int i = 0; i < n; i++)
    {
        for(int j = 0; j < n; j++)
            Console.Write(arr[i, j] + " ");
             
        Console.WriteLine();
    }
}
 
// Driver code
public static void Main(String[] args)
{
    int [,]arr = { { 1, 2, 3, 4 },
                   { 5, 6, 7, 8 },
                   { 9, 10, 11, 12 },
                   { 13, 14, 15, 16 } };
    rotate(arr);
    printMatrix(arr);
}
}
 
// This code is contributed by aashish1995

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