右转K次后打印数组|设置 2

给定一个大小为N的数组arr[]和一个值K，任务是打印向右旋转K次的数组。

例子：

Input: arr = {1, 3, 5, 7, 9}, K = 2
Output: 7 9 1 3 5

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

算法：给定的问题可以通过反转子数组来解决。可以按照以下步骤解决问题：

将所有数组元素从1反转为N -1
将数组元素从1反转为K – 1
将数组元素从K反转为N -1

C++

// C++ implementation for the above approach
 
#include 
using namespace std;
 
// Function to rotate the array
// to the right, K times
void RightRotate(int Array[], int N, int K)
{
 
    // Reverse all the array elements
    reverse(Array, Array + N);
 
    // Reverse the first k elements
    reverse(Array, Array + K);
 
    // Reverse the elements from K
    // till the end of the array
    reverse(Array + K, Array + N);
 
    // Print the array after rotation
    for (int i = 0; i < N; i++) {
 
        cout << Array[i] << " ";
    }
 
    cout << endl;
}
 
// Driver code
int main()
{
 
    // Initialize the array
    int Array[] = { 1, 2, 3, 4, 5 };
 
    // Find the size of the array
    int N = sizeof(Array) / sizeof(Array[0]);
 
    // Initialize K
    int K = 4;
 
    // Call the function and
    // print the answer
    RightRotate(Array, N, K);
 
    return 0;
}

Java

/*package whatever //do not write package name here */
import java.io.*;
class GFG
{
   
    // Function to rotate the array
    // to the right, K times
    static void RightRotate(int[] Array, int N, int K)
    {
 
        // Reverse all the array elements
        for (int i = 0; i < N / 2; i++) {
            int temp = Array[i];
            Array[i] = Array[N - i - 1];
            Array[N - i - 1] = temp;
        }
 
        // Reverse the first k elements
        for (int i = 0; i < K / 2; i++) {
            int temp = Array[i];
            Array[i] = Array[K - i - 1];
            Array[K - i - 1] = temp;
        }
 
        // Reverse the elements from K
        // till the end of the array
        for (int i = 0; i < (K + N) / 2; i++) {
            int temp = Array[(i + K) % N];
            Array[(i + K) % N] = Array[(N - i + K - 1) % N];
            Array[(N - i + K - 1) % N] = temp;
        }
 
        // Print the array after rotation
        for (int i = 0; i < N; i++) {
 
            System.out.print(Array[i] + " ");
        }
 
        System.out.println();
    }
 
    // Driver code
    public static void main(String[] args)
    {
       
        // Initialize the array
        int Array[] = { 1, 2, 3, 4, 5 };
 
        // Find the size of the array
        int N = Array.length;
 
        // Initialize K
        int K = 4;
 
        // Call the function and
        // print the answer
        RightRotate(Array, N, K);
    }
}
 
// This code is contributed by maddler.

Python3

# Python program for the above approach
import math
 
# Function to rotate the array
# to the right, K times
def RightRotate(Array, N, K):
 
    # Reverse all the array elements
    for i in range(math.ceil(N / 2)):
        temp = Array[i]
        Array[i] = Array[N - i - 1]
        Array[N - i - 1] = temp
 
    # Reverse the first k elements
    for i in range(math.ceil(K / 2)):
        temp = Array[i]
        Array[i] = Array[K - i - 1]
        Array[K - i - 1] = temp
 
    # Reverse the elements from K
    # till the end of the array
    for i in range(math.ceil((K + N) / 2)):
        temp = Array[(i + K) % N]
        Array[(i + K) % N] = Array[(N - i + K - 1) % N]
        Array[(N - i + K - 1) % N] = temp
 
    # Print the array after rotation
    for i in range(N):
        print(Array[i], end=" ")
 
 
# Driver Code
arr = [1, 2, 3, 4, 5]
N = len(arr)
K = 4
 
# Call the function and
# print the answer
RightRotate(arr, N, K)
 
# This code is contributed by Saurabh Jaiswal

C#

// C# program for the above approach
using System;
class GFG
{
    // Function to rotate the array
    // to the right, K times
    static void RightRotate(int []Array, int N, int K)
    {
 
        // Reverse all the array elements
        for (int i = 0; i < N / 2; i++) {
            int temp = Array[i];
            Array[i] = Array[N - i - 1];
            Array[N - i - 1] = temp;
        }
 
        // Reverse the first k elements
        for (int i = 0; i < K / 2; i++) {
            int temp = Array[i];
            Array[i] = Array[K - i - 1];
            Array[K - i - 1] = temp;
        }
 
        // Reverse the elements from K
        // till the end of the array
        for (int i = 0; i < (K + N) / 2; i++) {
            int temp = Array[(i + K) % N];
            Array[(i + K) % N] = Array[(N - i + K - 1) % N];
            Array[(N - i + K - 1) % N] = temp;
        }
 
        // Print the array after rotation
        for (int i = 0; i < N; i++) {
 
            Console.Write(Array[i] + " ");
        }
    }
 
    // Driver code
    public static void Main()
    {
       
        // Initialize the array
        int []Array = { 1, 2, 3, 4, 5 };
 
        // Find the size of the array
        int N = Array.Length;
 
        // Initialize K
        int K = 4;
 
        // Call the function and
        // print the answer
        RightRotate(Array, N, K);
    }
}
// This code is contributed by Samim Hossain Mondal.

Javascript

输出

2 3 4 5 1

时间复杂度：O(N)
辅助空间：O(1)