右旋转K次后打印阵列

给定大小为N的数组和值K ，我们需要围绕该数组右旋转数组。如何快速打印正确的旋转阵列?

例子 ：

Input: Array[] = {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: Array[] = {1, 2, 3, 4, 5}, K = 4.
Output: 2 3 4 5 1      

方法：

1. 我们首先将K的模数乘以N(K = K％N)，因为每旋转N圈后数组将与初始数组相同。
2. 现在，我们将数组从i = 0迭代到i = N-1并检查，
   • 如果i ，则打印最右边的第K个元素(a [N + i -K])。除此以外，
   • 在“ K”个元素(a [i – K])之后打印数组。

下面是上述方法的实现。

C++

// C++ implementation of right rotation 
// of an array K number of times
#include
using namespace std;
  
// Function to rightRotate array
void RightRotate(int a[], int n, int k)
{
      
    // If rotation is greater 
    // than size of array
    k = k % n;
  
    for(int i = 0; i < n; i++)
    {
       if(i < k)
       {
             
           // Printing rightmost 
           // kth elements
           cout << a[n + i - k] << " ";
       }
       else
       {
             
           // Prints array after
           // 'k' elements
           cout << (a[i - k]) << " ";
       }
    }
    cout << "\n";
}
      
// Driver code
int main()
{
    int Array[] = { 1, 2, 3, 4, 5 };
    int N = sizeof(Array) / sizeof(Array[0]);
    int K = 2;
      
    RightRotate(Array, N, K);
}
  
// This code is contributed by Surendra_Gangwar

Java

// Java Implementation of Right Rotation 
// of an Array K number of times
import java.util.*;
import java.lang.*;
import java.io.*;
  
class Array_Rotation
{
  
// Function to rightRotate array
static void RightRotate(int a[], 
                        int n, int k)
{
      
    // If rotation is greater 
    // than size of array
    k=k%n;
  
    for(int i = 0; i < n; i++)
    {
        if(i

---

Python3

# Python3 implementation of right rotation 
# of an array K number of times
  
# Function to rightRotate array
def RightRotate(a, n, k):
  
    # If rotation is greater 
    # than size of array
    k = k % n;
  
    for i in range(0, n):
  
        if(i < k):
  
            # Printing rightmost 
            # kth elements
            print(a[n + i - k], end = " ");
  
        else:
  
            # Prints array after
            # 'k' elements
            print(a[i - k], end = " ");
  
    print("\n");
  
# Driver code
Array = [ 1, 2, 3, 4, 5 ];
N = len(Array);
K = 2;
      
RightRotate(Array, N, K);
  
# This code is contributed by Code_Mech

C#

// C# implementation of right rotation 
// of an array K number of times
using System;
class GFG{
  
// Function to rightRotate array
static void RightRotate(int []a, 
                        int n, int k)
{
  
    // If rotation is greater 
    // than size of array
    k = k % n;
  
    for(int i = 0; i < n; i++)
    {
       if(i < k)
       {
             
           // Printing rightmost 
           // kth elements
           Console.Write(a[n + i - k] + " ");
       }
       else
       {
             
           // Prints array after
           // 'k' elements
           Console.Write(a[i - k] + " ");
       }
    }
    Console.WriteLine();
}
      
// Driver code
public static void Main(String []args)
{
    int []Array = { 1, 2, 3, 4, 5 };
    int N = Array.Length;
    int K = 2;
      
    RightRotate(Array, N, K);
}
}
  
// This code is contributed by Rohit_ranjan

输出：

4 5 1 2 3

时间复杂度：O(n)
辅助空间：O(1)