将矩阵旋转 180 度的Java程序

给定一个方阵，任务是我们在不使用任何额外空间的情况下将其逆时针方向旋转 180 度。

例子 ：

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

Input :  1 2 3 4 
         5 6 7 8 
         9 0 1 2 
         3 4 5 6 
Output : 6 5 4 3 
         2 1 0 9 
         8 7 6 5 
         4 3 2 1

方法：1（仅打印旋转矩阵）
这个问题的解决方案是将矩阵旋转 180 度，我们可以很容易地遵循该步骤

Matrix =  a00 a01 a02
          a10 a11 a12
          a20 a21 a22

when we rotate it by 90 degree
then matrix is
Matrix = a02 a12 a22
         a01 a11 a21
         a00 a10 a20
  
when we rotate it by again 90 
degree then matrix is 
Matrix = a22 a21 a20
         a12 a11 a10
         a02 a01 a00 

从上图中，我们可以简单地将矩阵旋转 180 度，然后我们将不得不以相反的方式打印给定的矩阵。

Java

// Java program to rotate a
// matrix by 180 degrees
import java.util.*;
  
class GFG {
    static int N = 3;
  
    // Function to Rotate the
    // matrix by 180 degree
    static void rotateMatrix(int mat[][])
    {
  
        // Simply print from last
        // cell to first cell.
        for (int i = N - 1; i >= 0; i--) {
            for (int j = N - 1; j >= 0; j--)
                System.out.print(mat[i][j] + " ");
  
            System.out.println();
        }
    }
  
    // Driver Code
    public static void main(String[] args)
    {
        int[][] mat = { { 1, 2, 3 },
                        { 4, 5, 6 },
                        { 7, 8, 9 } };
  
        rotateMatrix(mat);
    }
}
  
// This code is contributed by ChitraNayal

输出 ：

9 8 7 
 6 5 4 
 3 2 1  

时间复杂度： O(N*N)
辅助空间： O(1)

方法：2（就地旋转）
有四个步骤：
1-查找矩阵的转置。
2-反转转置的列。
3-找到矩阵的转置。
4- 转置的反转列

Let the given matrix be
1  2  3  4
5  6  7  8
9  10 11 12
13 14 15 16

First we find transpose.
1 5 9 13
2 6 10 14
3 7 11 15
4 8 12 16

Then we reverse elements of every column.
4 8 12 16
3 7 11 15
2 6 10 14
1 5  9 13

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

Then we reverse elements of every column again
16 15 14 13 
12 11 10 9 
8 7 6 5 
4 3 2 1

Java

// Java program for left
// rotation of matrix by 180
import java.util.*;
  
class GFG {
    static int R = 4, C = 4, t = 0;
  
    // Function to rotate the
    // matrix by 180 degree
    static void reverseColumns(int arr[][])
    {
        for (int i = 0; i < C; i++) {
            for (int j = 0, k = C - 1; j < k; j++, k--) {
                t = arr[j][i];
                arr[j][i] = arr[k][i];
                arr[k][i] = t;
            }
        }
    }
  
    // Function for transpose of matrix
    static void transpose(int arr[][])
    {
        for (int i = 0; i < R; i++) {
            for (int j = i; j < C; j++) {
                t = arr[i][j];
                arr[i][j] = arr[j][i];
                arr[j][i] = t;
            }
        }
    }
  
    // Function for display the matrix
    static void printMatrix(int arr[][])
    {
        for (int i = 0; i < R; i++) {
            for (int j = 0; j < C; j++)
                System.out.print(arr[i][j] + " ");
            System.out.println();
        }
    }
  
    // Function to anticlockwise
    // rotate matrix by 180 degree
    static void rotate180(int arr[][])
    {
        transpose(arr);
        reverseColumns(arr);
        transpose(arr);
        reverseColumns(arr);
    }
  
    // 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 } };
  
        rotate180(arr);
        printMatrix(arr);
    }
}
  
// This code is contributed by ChitraNayal

输出 ：

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

时间复杂度： O(R*C)
辅助空间： O(1)
在上面的代码中，矩阵的转置必须被找到两次，并且列必须被反转两次。
所以，我们可以有更好的解决方案。

方法：3（位置交换）
在这里，我们交换各个位置的值。

Java

public class GFG {
       /**
     * Reverse Row at specified index in the matrix
     * @param data matrix
     * @param index row index
     */
    private static void reverseRow(int[][] data, int index) {
        int cols = data[index].length;
        for (int i = 0; i < cols / 2; i++) {
            int temp = data[index][i];
            data[index][i] = data[index][cols - i - 1];
            data[index][cols - i - 1] = temp;
        }
    }
  
    /**
     * Print Matrix data
     * @param data matrix
     */
    private static void printMatrix(int[][] data) {
        for (int i = 0; i < data.length; i++) {
            for (int j = 0; j < data[i].length; j++) {
                System.out.print(data[i][j] + " ");
            }
            System.out.println("");
        }
    }
  
    /**
     * Rotate Matrix by 180 degrees
     * @param data matrix
     */
    private static void rotateMatrix180(int[][] data) {
        int rows = data.length;
        int cols = data[0].length;
  
        if (rows % 2 != 0) {
            //If N is odd reverse the middle row in the matrix
            reverseRow(data, data.length / 2);
        }
  
        //Swap the value of matrix [i][j] with [rows - i - 1][cols - j - 1] for half the rows size. 
        for (int i = 0; i <= (rows/2) - 1; i++) {
            for (int j = 0; j < cols; j++) {
                int temp = data[i][j];
                data[i][j] = data[rows - i - 1][cols - j - 1];
                data[rows - i - 1][cols - j - 1] = temp;
            }
        }
    }
  
    public static void main(String[] args) {
        int[][] data = {
                {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}
        };
  
        //Rotate Matrix
        rotateMatrix180(data);
  
        //Print Matrix
        printMatrix(data);
    }
}

输出 ：

25 24 23 22 21 
20 19 18 17 16 
15 14 13 12 11 
10 9 8 7 6 
5 4 3 2 1 

时间复杂度： O(R*C)
辅助空间： O(1)

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