给定一个方阵，将其逆时针旋转90度，不使用任何额外空间。
null
例如：
Input: 1 2 3 4 5 6 7 8 9Output: 3 6 9 2 5 8 1 4 7 Rotated the input matrix by90 degrees in anti-clockwise direction.Input: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Output: 4 8 12 16 3 7 11 15 2 6 10 14 1 5 9 13Rotated the input matrix by90 degrees in anti-clockwise direction.
另一篇文章中已经讨论了一种需要额外空间的方法： 原地旋转方阵90度|设置1
这篇文章用一种空间优化的不同方法讨论了同样的问题。 方法： 其思想是找到矩阵的转置，然后反转转置矩阵的列。 下面是一个例子来说明这是如何工作的。
![图片[1]-在不使用任何额外空间的情况下将矩阵旋转90度|集2-yiteyi-C++库](https://www.yiteyi.com/wp-content/uploads/geeks/20200407/geeks_untitled215.png)
算法：
- 要解决给定的问题，有两项任务。第一个是找到转置，第二个是在不使用额外空间的情况下反转列
- 矩阵的转置是指矩阵在其对角线上翻转，即元素的行索引变为列索引，反之亦然。因此，为了找到转置，将（i，j）位置的元素与（j，i）交换。运行两个循环，外部循环从0到行计数，内部循环从0到外部循环的索引。
- 为了反转转置矩阵的列，运行两个嵌套循环，外部循环从0到列计数，内部循环从0到行计数/2，将（i，j）处的元素与（i，row[count-1-j]）交换，其中i和j分别是内部循环和外部循环的索引。
C++
// C++ program for left // rotation of matrix by 90 degree // without using extra space #include <bits/stdc++.h> using namespace std; #define R 4 #define C 4 // After transpose we swap // elements of column // one by one for finding left // rotation of matrix // by 90 degree void reverseColumns( int arr[R][C]) { for ( int i = 0; i < C; i++) for ( int j = 0, k = C - 1; j < k; j++, k--) swap(arr[j][i], arr[k][i]); } // Function for do transpose of matrix void transpose( int arr[R][C]) { for ( int i = 0; i < R; i++) for ( int j = i; j < C; j++) swap(arr[i][j], arr[j][i]); } // Function for print matrix void printMatrix( int arr[R][C]) { for ( int i = 0; i < R; i++) { for ( int j = 0; j < C; j++) cout << arr[i][j] << " " ; cout << '' ; } } // Function to anticlockwise // rotate matrix by 90 degree void rotate90( int arr[R][C]) { transpose(arr); reverseColumns(arr); } // Driven code int main() { int arr[R][C] = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 }, { 13, 14, 15, 16 } }; rotate90(arr); printMatrix(arr); return 0; } |
JAVA
// JAVA Code for left Rotation of a // matrix by 90 degree without using // any extra space import java.util.*; class GFG { // After transpose we swap elements of // column one by one for finding left // rotation of matrix by 90 degree static void reverseColumns( int arr[][]) { for ( int i = 0 ; i < arr[ 0 ].length; i++) for ( int j = 0 , k = arr[ 0 ].length - 1 ; j < k; j++, k--) { int temp = arr[j][i]; arr[j][i] = arr[k][i]; arr[k][i] = temp; } } // Function for do transpose of matrix static void transpose( int arr[][]) { for ( int i = 0 ; i < arr.length; i++) for ( int j = i; j < arr[ 0 ].length; j++) { int temp = arr[j][i]; arr[j][i] = arr[i][j]; arr[i][j] = temp; } } // Function for print matrix static void printMatrix( int arr[][]) { for ( int i = 0 ; i < arr.length; i++) { for ( int j = 0 ; j < arr[ 0 ].length; j++) System.out.print(arr[i][j] + " " ); System.out.println( "" ); } } // Function to anticlockwise rotate // matrix by 90 degree static void rotate90( int arr[][]) { transpose(arr); reverseColumns(arr); } /* Driver program to test above function */ public static void main(String[] args) { int arr[][] = { { 1 , 2 , 3 , 4 }, { 5 , 6 , 7 , 8 }, { 9 , 10 , 11 , 12 }, { 13 , 14 , 15 , 16 } }; rotate90(arr); printMatrix(arr); } } // This code is contributed by Arnav Kr. Mandal. |
C#
// C# program for left rotation // of matrix by 90 degree // without using extra space using System; class GFG { static int R = 4; static int C = 4; // After transpose we swap // elements of column one // by one for finding left // rotation of matrix by // 90 degree static void reverseColumns( int [, ] arr) { for ( int i = 0; i < C; i++) for ( int j = 0, k = C - 1; j < k; j++, k--) { int temp = arr[j, i]; arr[j, i] = arr[k, i]; arr[k, i] = temp; } } // Function for do // transpose of matrix static void transpose( int [, ] arr) { for ( int i = 0; i < R; i++) for ( int j = i; j < C; j++) { int temp = arr[j, i]; arr[j, i] = arr[i, j]; arr[i, j] = temp; } } // Function for print matrix static void printMatrix( int [, ] arr) { for ( int i = 0; i < R; i++) { for ( int j = 0; j < C; j++) Console.Write(arr[i, j] + " " ); Console.WriteLine( "" ); } } // Function to anticlockwise // rotate matrix by 90 degree static void rotate90( int [, ] arr) { transpose(arr); reverseColumns(arr); } // Driver code static void Main() { int [, ] arr = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 }, { 13, 14, 15, 16 } }; rotate90(arr); printMatrix(arr); } // This code is contributed // by Sam007 } |
Python3
# Python 3 program for left rotation of matrix by 90 # degree without using extra space R = 4 C = 4 # After transpose we swap elements of column # one by one for finding left rotation of matrix # by 90 degree def reverseColumns(arr): for i in range (C): j = 0 k = C - 1 while j < k: t = arr[j][i] arr[j][i] = arr[k][i] arr[k][i] = t j + = 1 k - = 1 # Function for do transpose of matrix def transpose(arr): for i in range (R): for j in range (i, C): t = arr[i][j] arr[i][j] = arr[j][i] arr[j][i] = t # Function for print matrix def printMatrix(arr): for i in range (R): for j in range (C): print ( str (arr[i][j]), end = " " ) print () # Function to anticlockwise rotate matrix # by 90 degree def rotate90(arr): transpose(arr) reverseColumns(arr) # Driven code arr = [[ 1 , 2 , 3 , 4 ], [ 5 , 6 , 7 , 8 ], [ 9 , 10 , 11 , 12 ], [ 13 , 14 , 15 , 16 ] ] rotate90(arr) printMatrix(arr) |
PHP
<?php // PHP program for left rotation of matrix by 90 $R = 4; $C = 4; // Function to rotate the matrix by 90 degree function reverseColumns(& $arr ) { global $C ; for ( $i = 0; $i < $C ; $i ++) { for ( $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 function transpose(& $arr ) { global $R , $C ; for ( $i = 0; $i < $R ; $i ++) { for ( $j = $i ; $j < $C ; $j ++) { $t = $arr [ $i ][ $j ]; $arr [ $i ][ $j ] = $arr [ $j ][ $i ]; $arr [ $j ][ $i ] = $t ; } } } // Function for display the matrix function printMatrix(& $arr ) { global $R , $C ; for ( $i = 0; $i < $R ; $i ++) { for ( $j = 0; $j < $C ; $j ++) echo $arr [ $i ][ $j ]. " " ; echo "" ; } } // Function to anticlockwise rotate matrix // by 90 degree function rotate90(& $arr ) { transpose( $arr ); reverseColumns( $arr ); } // Driven code $arr = array ( array ( 1, 2, 3, 4 ), array ( 5, 6, 7, 8 ), array ( 9, 10, 11, 12 ), array ( 13, 14, 15, 16 ) ); rotate90( $arr ); printMatrix( $arr ); return 0; ?> |
Javascript
<script> // JAVASCRIPT Code for left Rotation of a // matrix by 90 degree without using // any extra space // After transpose we swap elements of // column one by one for finding left // rotation of matrix by 90 degree function reverseColumns(arr) { for (let i = 0; i < arr[0].length; i++) for (let j = 0, k = arr[0].length - 1; j < k; j++, k--) { let temp = arr[j][i]; arr[j][i] = arr[k][i]; arr[k][i] = temp; } } // Function for do transpose of matrix function transpose(arr) { for (let i = 0; i < arr.length; i++) for (let j = i; j < arr[0].length; j++) { let temp = arr[j][i]; arr[j][i] = arr[i][j]; arr[i][j] = temp; } } // Function for print matrix function printMatrix(arr) { for (let i = 0; i < arr.length; i++) { for (let j = 0; j < arr[0].length; j++) document.write(arr[i][j] + " " ); document.write( "<br>" ); } } // Function to anticlockwise rotate // matrix by 90 degree function rotate90(arr) { transpose(arr); reverseColumns(arr); } /* Driver program to test above function */ let arr = [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16] ]; rotate90(arr) printMatrix(arr) // This code is contributed by rag2127 </script> |
输出
4 8 12 16 3 7 11 15 2 6 10 14 1 5 9 13
复杂性分析：
- 时间复杂性： O（R*C）。 矩阵被遍历两次，因此复杂度为O（R*C）。
- 空间复杂性： O（1）。 空间复杂度是恒定的，因为不需要额外的空间。
另一种方法是使用 矩阵的单次遍历：
这个想法是沿着矩阵的边界移动元素的位置 0 每个边界的逆时针方向。矩阵中存在这样的（n/2-1）边界。
算法：
- 迭代矩阵中的所有边界。共有（n/2-1）个边界
- 对于每个边界，取4个角元素并交换它们，使4个角元素逆时针旋转。然后沿着边缘（左、右、上、下）取下4个元素，并逆时针交换它们。只要该特定边界中的所有元素都以90度旋转，就可以继续 0 逆时针方向。
- 然后移动到下一个内边界，继续这个过程，只要整个矩阵旋转90度 0 逆时针方向。
![图片[2]-在不使用任何额外空间的情况下将矩阵旋转90度|集2-yiteyi-C++库](https://www.yiteyi.com/wp-content/uploads/geeks/20210618/geeks_gfgdraw1-300x184.png)
原始数组
![图片[3]-在不使用任何额外空间的情况下将矩阵旋转90度|集2-yiteyi-C++库](https://www.yiteyi.com/wp-content/uploads/geeks/20210618/geeks_gfgdraw2png-300x178.png)
总n/2-1边界
![图片[4]-在不使用任何额外空间的情况下将矩阵旋转90度|集2-yiteyi-C++库](https://www.yiteyi.com/wp-content/uploads/geeks/20210618/geeks_gfgdraw7.png)
最外面的边界
![图片[5]-在不使用任何额外空间的情况下将矩阵旋转90度|集2-yiteyi-C++库](https://www.yiteyi.com/wp-content/uploads/geeks/20210618/geeks_gfgdraw6-300x179.png)
下一个内边界
C++14
// C++ program for left rotation of matrix // by 90 degree without using extra space #include <bits/stdc++.h> using namespace std; #define R 4 #define C 4 // Function to rotate matrix anticlockwise by 90 degrees. void rotateby90( int arr[R][C]) { int n = R; // n=size of the square matrix int a = 0, b = 0, c = 0, d = 0; // iterate over all the boundaries of the matrix for ( int i = 0; i <= n / 2 - 1; i++) { // for each boundary, keep on taking 4 elements (one // each along the 4 edges) and swap them in // anticlockwise manner for ( int j = 0; j <= n - 2 * i - 2; j++) { a = arr[i + j][i]; b = arr[n - 1 - i][i + j]; c = arr[n - 1 - i - j][n - 1 - i]; d = arr[i][n - 1 - i - j]; arr[i + j][i] = d; arr[n - 1 - i][i + j] = a; arr[n - 1 - i - j][n - 1 - i] = b; arr[i][n - 1 - i - j] = c; } } } // Function for print matrix void printMatrix( int arr[R][C]) { for ( int i = 0; i < R; i++) { for ( int j = 0; j < C; j++) cout << arr[i][j] << " " ; cout << '' ; } } // Driven code int main() { int arr[R][C] = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 }, { 13, 14, 15, 16 } }; rotateby90(arr); printMatrix(arr); return 0; } // This code is contributed by Md. Enjamum // Hossain(enja_2001) |
JAVA
// JAVA Code for left Rotation of a matrix // by 90 degree without using any extra space import java.util.*; class GFG { // Function to rotate matrix anticlockwise by 90 // degrees. static void rotateby90( int arr[][]) { int n = arr.length; int a = 0 , b = 0 , c = 0 , d = 0 ; // iterate over all the boundaries of the matrix for ( int i = 0 ; i <= n / 2 - 1 ; i++) { // for each boundary, keep on taking 4 elements // (one each along the 4 edges) and swap them in // anticlockwise manner for ( int j = 0 ; j <= n - 2 * i - 2 ; j++) { a = arr[i + j][i]; b = arr[n - 1 - i][i + j]; c = arr[n - 1 - i - j][n - 1 - i]; d = arr[i][n - 1 - i - j]; arr[i + j][i] = d; arr[n - 1 - i][i + j] = a; arr[n - 1 - i - j][n - 1 - i] = b; arr[i][n - 1 - i - j] = c; } } } // Function for print matrix static void printMatrix( int arr[][]) { for ( int i = 0 ; i < arr.length; i++) { for ( int j = 0 ; j < arr[ 0 ].length; j++) System.out.print(arr[i][j] + " " ); System.out.println( "" ); } } /* Driver program to test above function */ public static void main(String[] args) { int arr[][] = { { 1 , 2 , 3 , 4 }, { 5 , 6 , 7 , 8 }, { 9 , 10 , 11 , 12 }, { 13 , 14 , 15 , 16 } }; rotateby90(arr); printMatrix(arr); } } // This code is contributed by Md. Enjamum // Hossain(enja_2001) |
C#
// C# Code for left Rotation of a matrix // by 90 degree without using any extra space using System; using System.Collections.Generic; public class GFG { // Function to rotate matrix anticlockwise by 90 // degrees. static void rotateby90( int [,]arr) { int n = arr.GetLength(0); int a = 0, b = 0, c = 0, d = 0; // iterate over all the boundaries of the matrix for ( int i = 0; i <= n / 2 - 1; i++) { // for each boundary, keep on taking 4 elements // (one each along the 4 edges) and swap them in // anticlockwise manner for ( int j = 0; j <= n - 2 * i - 2; j++) { a = arr[i + j,i]; b = arr[n - 1 - i,i + j]; c = arr[n - 1 - i - j,n - 1 - i]; d = arr[i,n - 1 - i - j]; arr[i + j,i] = d; arr[n - 1 - i,i + j] = a; arr[n - 1 - i - j,n - 1 - i] = b; arr[i,n - 1 - i - j] = c; } } } // Function for print matrix static void printMatrix( int [,]arr) { for ( int i = 0; i < arr.GetLength(0); i++) { for ( int j = 0; j < arr.GetLength(1); j++) Console.Write(arr[i,j] + " " ); Console.WriteLine( "" ); } } /* Driver program to test above function */ public static void Main(String[] args) { int [,]arr = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 }, { 13, 14, 15, 16 } }; rotateby90(arr); printMatrix(arr); } } // This code is contributed by umadevi9616 |
Javascript
<script> // JavaScript Code for left Rotation of a matrix // by 90 degree without using any extra space // Function to rotate matrix anticlockwise by 90 // degrees. function rotateby90(arr) { let n = arr.length; let a = 0, b = 0, c = 0, d = 0; // iterate over all the boundaries of the matrix for (let i = 0; i <= n / 2 - 1; i++) { // for each boundary, keep on taking 4 elements // (one each along the 4 edges) and swap them in // anticlockwise manner for (let j = 0; j <= n - 2 * i - 2; j++) { a = arr[i + j][i]; b = arr[n - 1 - i][i + j]; c = arr[n - 1 - i - j][n - 1 - i]; d = arr[i][n - 1 - i - j]; arr[i + j][i] = d; arr[n - 1 - i][i + j] = a; arr[n - 1 - i - j][n - 1 - i] = b; arr[i][n - 1 - i - j] = c; } } } // Function for print matrix function printMatrix(arr) { for (let i = 0; i < arr.length; i++) { for (let j = 0; j < arr[0].length; j++) document.write(arr[i][j] + " " ); document.write( "<br>" ); } } /* Driver program to test above function */ let arr=[[ 1, 2, 3, 4 ], [ 5, 6, 7, 8 ], [ 9, 10, 11, 12 ], [ 13, 14, 15, 16 ]]; rotateby90(arr); printMatrix(arr); // This code is contributed by avanitrachhadiya2155 </script> |
输出
4 8 12 16 3 7 11 15 2 6 10 14 1 5 9 13
复杂性分析：
时间复杂性 ：O（R*C）。 矩阵只遍历一次，因此时间复杂度为O（R*C）。
空间复杂性：O（1） .由于不需要额外的空间，因此空间复杂度为O（1）。
实施 ：让我们来看一个解决问题的方法 Python numpy 这可以用来得出特定的解决方案。
Python3
# Alternative implementation using numpy import numpy # Driven code arr = [[ 1 , 2 , 3 , 4 ], [ 5 , 6 , 7 , 8 ], [ 9 , 10 , 11 , 12 ], [ 13 , 14 , 15 , 16 ] ] # Define flip algorithm (Note numpy.flip is a builtin f # function for versions > v.1.12.0) def flip(m, axis): if not hasattr (m, 'ndim' ): m = asarray(m) indexer = [ slice ( None )] * m.ndim try : indexer[axis] = slice ( None , None , - 1 ) except IndexError: raise ValueError( "axis =% i is invalid for the % i-dimensional input array" % (axis, m.ndim)) return m[ tuple (indexer)] # Transpose the matrix trans = numpy.transpose(arr) # Flip the matrix anti-clockwise (1 for clockwise) flipmat = flip(trans, 0 ) print ( "numpy implementation" ) print (flipmat) |
输出：
numpy implementation[[ 4 8 12 16] [ 3 7 11 15] [ 2 6 10 14] [ 1 5 9 13]]
注： 上述步骤/程序会向左（或逆时针）旋转。让我们看看如何做正确的旋转或顺时针旋转。方法也会类似。找到矩阵的转置，然后反转转置矩阵的行。 就是这样做的。
![图片[6]-在不使用任何额外空间的情况下将矩阵旋转90度|集2-yiteyi-C++库](https://www.yiteyi.com/wp-content/uploads/geeks/20200407/geeks_untitled312.png)
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沿着边界旋转
我们可以从给定矩阵的前4个角开始，然后不断增加行和列索引，以进行移动。
在任何给定的时刻，我们将有四个角lu（左上）、ld（左下）、ru（右上）、rd（右下）。
要向左旋转，我们首先交换ru和ld，然后交换lu和ld，最后交换ru和rd。
| 汝 | 鲁 | |
| 研发部 | ld |
C++
#include <bits/stdc++.h> using namespace std; //Function to rotate matrix anticlockwise by 90 degrees. void rotateby90(vector<vector< int > >& matrix) { int n=matrix.size(); int mid; if (n%2==0) mid=n/2-1; else mid=n/2; for ( int i=0,j=n-1;i<=mid;i++,j--) { for ( int k=0;k<j-i;k++) { swap(matrix[i][j-k],matrix[j][i+k]); //ru ld swap(matrix[i+k][i],matrix[j][i+k]); //lu ld swap(matrix[i][j-k],matrix[j-k][j]); //ru rd } } } void printMatrix(vector<vector< int >>& arr) { int n=arr.size(); for ( int i=0;i<n;i++) { for ( int j=0;j<n;j++) { cout<<arr[i][j]<< " " ; } cout<<endl; } } int main() { vector<vector< int >> arr = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 }, { 13, 14, 15, 16 } }; rotateby90(arr); printMatrix(arr); return 0; } |
JAVA
import java.util.*; class GFG{ private static int [][] swap( int [][] matrix, int x1, int x2, int y1, int y2) { int temp = matrix[x1][x2] ; matrix[x1][x2] = matrix[y1][y2]; matrix[y1][y2] = temp; return matrix; } //Function to rotate matrix anticlockwise by 90 degrees. static int [][] rotateby90( int [][] matrix) { int n=matrix.length; int mid; if (n % 2 == 0 ) mid = n / 2 - 1 ; else mid = n / 2 ; for ( int i = 0 , j = n - 1 ; i <= mid; i++, j--) { for ( int k = 0 ; k < j - i; k++) { matrix= swap(matrix, i, j - k, j, i + k); //ru ld matrix= swap(matrix, i + k, i, j, i + k); //lu ld matrix= swap(matrix, i, j - k, j - k, j); //ru rd } } return 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(); } } public static void main(String[] args) { int [][] arr = { { 1 , 2 , 3 , 4 }, { 5 , 6 , 7 , 8 }, { 9 , 10 , 11 , 12 }, { 13 , 14 , 15 , 16 } }; arr = rotateby90(arr); printMatrix(arr); } } // This code is contributed by umadevi9616 |
C#
using System; public class GFG{ private static int [,] swap( int [,] matrix, int x1, int x2, int y1, int y2) { int temp = matrix[x1,x2] ; matrix[x1,x2] = matrix[y1,y2]; matrix[y1,y2] = temp; return matrix; } //Function to rotate matrix anticlockwise by 90 degrees. static int [,] rotateby90( int [,] matrix) { int n=matrix.GetLength(0); int mid; if (n % 2 == 0) mid = (n / 2 - 1); else mid = (n / 2); for ( int i = 0, j = n - 1; i <= mid; i++, j--) { for ( int k = 0; k < j - i; k++) { matrix= swap(matrix, i, j - k, j, i + k); //ru ld matrix= swap(matrix, i + k, i, j, i + k); //lu ld matrix= swap(matrix, i, j - k, j - k, j); //ru rd } } return 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(); } } public static void Main(String[] args) { int [,] arr = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 }, { 13, 14, 15, 16 } }; arr = rotateby90(arr); printMatrix(arr); } } // This code is contributed by Rajput-Ji |
Javascript
<script> function swap(matrix , x1 , x2 , y1 , y2) { var temp = matrix[x1][x2]; matrix[x1][x2] = matrix[y1][y2]; matrix[y1][y2] = temp; return matrix; } // Function to rotate matrix anticlockwise by 90 degrees. function rotateby90(matrix) { var n = matrix.length; var mid; if (n % 2 == 0) mid = n / 2 - 1; else mid = n / 2; for (i = 0, j = n - 1; i <= mid; i++, j--) { for (k = 0; k < j - i; k++) { matrix = swap(matrix, i, j - k, j, i + k); // ru ld matrix = swap(matrix, i + k, i, j, i + k); // lu ld matrix = swap(matrix, i, j - k, j - k, j); // ru rd } } return matrix; } function printMatrix(arr) { var n = arr.length; for (i = 0; i < n; i++) { for (j = 0; j < n; j++) { 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 ] ]; arr = rotateby90(arr); printMatrix(arr); // This code is contributed by Rajput-Ji </script> |
输出：
4 8 12 16 3 7 11 15 2 6 10 14 1 5 9 13
时间复杂度：O（n^2）
空间复杂性：O（1）
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