使用指针向右旋转给定数组 K 次

给定一个大小为 N 的数组 arr[] 和一个整数 K，任务是将数组右转 K 次。

例子：

Input: arr[] = {1, 3, 5, 7, 9}, K = 2
Output: 7 9 1 3 5
Explanation: After 1st rotation – {9, 1, 3, 5, 7}
After 2nd rotation – {7, 9, 1, 3, 5}

Input: {1, 2, 3, 4, 5, 6}, K = 2
Output: 5 6 1 2 3 4

方法：这里讨论了基于反转部分数组的朴素方法和方法。

基于指针的方法：这个概念的基础是数组旋转的反转算法。数组分为两部分，其中第一部分的大小为 (NK)，末端部分的大小为 K。这两个部分分别反转。然后整个数组被反转。

下面是上述方法的实现：

C++

// C++ code to implement above approach
#include 
using namespace std;
 
// Function to print the array
void print(int arr[], int N)
{
    for (int i = 0; i < N; i++)
        cout << *(arr + i) << " ";
}
 
// Function to reverse the array
// from start to end index
void reverse(int arr[], int start, int end)
{
    int temp;
    int size = end - start;
 
    // Reversal based on pointer approach
    for (int i = 0; i < (size / 2); i++) {
        temp = *(arr + i + start);
        *(arr + i + start) = *(arr + start
                             + size - i - 1);
        *(arr + start + size - i - 1) = temp;
    }
}
 
// Function to right rotate the array K times
void right(int arr[], int K, int N)
{
    reverse(arr, 0, N - K);
    reverse(arr, N - K, N);
    reverse(arr, 0, N);
    print(arr, N);
}
 
// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5, 6 };
    int N = sizeof(arr) / sizeof(arr[0]);
    int K = 2;
    right(arr, K, N);
    return 0;
}

Java

// Java code to implement above approach
import java.util.*;
 
class GFG{
 
// Function to print the array
static void print(int arr[], int N)
{
    for (int i = 0; i < N; i++)
        System.out.print(arr[i]+ " ");
}
 
// Function to reverse the array
// from start to end index
static int[] reverse(int arr[], int start, int end)
{
    int temp;
    int size = end - start;
 
    // Reversal based on pointer approach
    for (int i = 0; i < (size / 2); i++) {
        temp = arr[ i + start];
        arr[i + start] = arr[start
                             + size - i - 1];
        arr[start + size - i - 1] = temp;
    }
    return arr;
}
 
// Function to right rotate the array K times
static void right(int arr[], int K, int N)
{
    arr = reverse(arr, 0, N - K);
    arr = reverse(arr, N - K, N);
    arr = reverse(arr, 0, N);
    print(arr, N);
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 3, 4, 5, 6 };
    int N = arr.length;
    int K = 2;
    right(arr, K, N);
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python code to implement above approach
 
# Function to print array
def print1(arr, N):
    for i in range(N):
        print(arr[i], end = " ");
 
# Function to reverse the array
# from start to end index
def reverse(arr, start, end):
    temp = 0;
    size = end - start;
 
    # Reversal based on pointer approach
    for i in range(size//2):
        temp = arr[i + start];
        arr[i + start] = arr[start + size - i - 1];
        arr[start + size - i - 1] = temp;
 
    return arr;
 
# Function to right rotate the array K times
def right(arr, K, N):
    arr = reverse(arr, 0, N - K);
    arr = reverse(arr, N - K, N);
    arr = reverse(arr, 0, N);
    print1(arr, N);
 
# Driver code
if __name__ == '__main__':
    arr = [1, 2, 3, 4, 5, 6];
    N = len(arr);
    K = 2;
    right(arr, K, N);
 
    # This code is contributed by 29AjayKumar

C#

// C# code to implement above approach
using System;
class GFG {
 
  // Function to print the array
  static void print(int[] arr, int N)
  {
    for (int i = 0; i < N; i++)
      Console.Write(arr[i] + " ");
  }
 
  // Function to reverse the array
  // from start to end index
  static void reverse(int[] arr, int start, int end)
  {
    int temp;
    int size = end - start;
 
    // Reversal based on pointer approach
    for (int i = 0; i < (size / 2); i++) {
      temp = arr[i + start];
      arr[i + start] = arr[start + size - i - 1];
      arr[start + size - i - 1] = temp;
    }
  }
 
  // Function to right rotate the array K times
  static void right(int[] arr, int K, int N)
  {
    reverse(arr, 0, N - K);
    reverse(arr, N - K, N);
    reverse(arr, 0, N);
    print(arr, N);
  }
 
  // Driver code
  public static void Main()
  {
    int[] arr = { 1, 2, 3, 4, 5, 6 };
    int N = arr.Length;
    int K = 2;
    right(arr, K, N);
  }
}
 
// This code is contributed by ukasp.

Javascript

    // JavaScript code to implement above approach
 
    // Function to print the array
    const print = (arr, N) => {
        for (let i = 0; i < N; i++)
            document.write(`${arr[i]} `);
    }
 
    // Function to reverse the array
    // from start to end index
    const reverse = (arr, start, end) => {
        let temp;
        let size = end - start;
 
        // Reversal based on pointer approach
        for (let i = 0; i < parseInt(size / 2); i++) {
            temp = arr[i + start];
            arr[i + start] = arr[start + size - i - 1];
            arr[start + size - i - 1] = temp;
        }
    }
 
    // Function to right rotate the array K times
    const right = (arr, K, N) => {
        reverse(arr, 0, N - K);
        reverse(arr, N - K, N);
        reverse(arr, 0, N);
        print(arr, N);
    }
 
    // Driver code
    let arr = [1, 2, 3, 4, 5, 6];
    let N = arr.length;
    let K = 2;
    right(arr, K, N);
 
// This code is contributed by rakeshsahni
 

输出

5 6 1 2 3 4 

时间复杂度： O(N)
辅助空间： O(1)