程序循环旋转一个数组 · GeeksForGeeks 中文系列教程

# 程序循环旋转一个数组 > 原文： [https://www.geeksforgeeks.org/c-program-cyclically-rotate-array-one/](https://www.geeksforgeeks.org/c-program-cyclically-rotate-array-one/) 给定一个数组，将数组顺时针方向顺时针旋转一个。 **示例**： ``` Input: arr[] = {1, 2, 3, 4, 5} Output: arr[] = {5, 1, 2, 3, 4} ``` 以下是步骤。 1）将最后一个元素存储在变量`x`中。 2）将所有元素向前移动一个位置。 3）将数组的第一个元素替换为`x`。 ## C++ ```cpp // C++ code for program // to cyclically rotate // an array by one # include <iostream> using namespace std; void rotate(int arr[], int n) { int x = arr[n - 1], i; for (i = n - 1; i > 0; i--) arr[i] = arr[i - 1]; arr[0] = x; } // Driver code int main() { int arr[] = {1, 2, 3, 4, 5}, i; int n = sizeof(arr) / sizeof(arr[0]); cout << "Given array is \n"; for (i = 0; i < n; i++) cout << arr[i]; rotate(arr, n); cout << "\nRotated array is\n"; for (i = 0; i < n; i++) cout << arr[i]; return 0; } // This code is contributed by jit_t ``` ## C ``` #include <stdio.h> void rotate(int arr[], int n) { int x = arr[n-1], i; for (i = n-1; i > 0; i--) arr[i] = arr[i-1]; arr[0] = x; } int main() { int arr[] = {1, 2, 3, 4, 5}, i; int n = sizeof(arr)/sizeof(arr[0]); printf("Given array is\n"); for (i = 0; i < n; i++) printf("%d ", arr[i]); rotate(arr, n); printf("\nRotated array is\n"); for (i = 0; i < n; i++) printf("%d ", arr[i]); return 0; } ``` ## Java ```java import java.util.Arrays; public class Test { static int arr[] = new int[]{1, 2, 3, 4, 5}; // Method for rotation static void rotate() { int x = arr[arr.length-1], i; for (i = arr.length-1; i > 0; i--) arr[i] = arr[i-1]; arr[0] = x; } /* Driver program */ public static void main(String[] args) { System.out.println("Given Array is"); System.out.println(Arrays.toString(arr)); rotate(); System.out.println("Rotated Array is"); System.out.println(Arrays.toString(arr)); } } ``` ## Python3 ```py # Python3 code for program to # cyclically rotate an array by one # Method for rotation def rotate(arr, n): x = arr[n - 1] for i in range(n - 1, 0, -1): arr[i] = arr[i - 1]; arr[0] = x; # Driver function arr= [1, 2, 3, 4, 5] n = len(arr) print ("Given array is") for i in range(0, n): print (arr[i], end = ' ') rotate(arr, n) print ("\nRotated array is") for i in range(0, n): print (arr[i], end = ' ') # This article is contributed # by saloni1297 ``` ## C# ```cs // C# code for program to cyclically // rotate an array by one using System; public class Test { static int []arr = new int[]{1, 2, 3, 4, 5}; // Method for rotation static void rotate() { int x = arr[arr.Length - 1], i; for (i = arr.Length - 1; i > 0; i--) arr[i] = arr[i-1]; arr[0] = x; } // Driver Code public static void Main() { Console.WriteLine("Given Array is"); string Original_array = string.Join(" ", arr); Console.WriteLine(Original_array); rotate(); Console.WriteLine("Rotated Array is"); string Rotated_array = string.Join(" ", arr); Console.WriteLine(Rotated_array); } } // This code is contributed by vt_m. ``` ## PHP ```php <?php // PHP code for program // to cyclically rotate // an array by one function rotate(&$arr, $n) { $x = $arr[$n - 1]; for ($i = $n - 1; $i > 0; $i--) $arr[$i] = $arr[$i - 1]; $arr[0] = $x; } // Driver code $arr = array(1, 2, 3, 4, 5); $n = sizeof($arr); echo "Given array is \n"; for ($i = 0; $i < $n; $i++) echo $arr[$i] . " "; rotate($arr, $n); echo "\nRotated array is\n"; for ($i = 0; $i < $n; $i++) echo $arr[$i] . " "; // This code is contributed // by ChitraNayal ?> ``` **输出**： ``` Given array is 1 2 3 4 5 Rotated array is 5 1 2 3 4 ``` 时间复杂度：`O(n)`当我们需要遍历所有元素时辅助空间：`O(1)` 通过使用[逆转算法](https://www.geeksforgeeks.org/program-for-array-rotation-continued-reversal-algorithm/)也可以解决上述问题如果您在上述代码/算法中发现任何错误，或者找到其他解决同一问题的方法，请发表评论