右旋K次后打印数组

给定一个大小为 N 的数组和一个值K ，我们需要围绕它右旋转数组。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 mod 乘以 N(K = K % N)，因为在每 N 次旋转后，数组将变得与初始数组相同。
2. 现在，我们将数组从 i = 0 迭代到 i = N-1 并检查，
   • 如果i < K ，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

Javascript

// Javascript implementation of right rotation
// of an array K number of times
 
// Function to rightRotate array
function RightRotate(a, n, k)
{
 
    // If rotation is greater
    // than size of array
    k = k % n;
 
    for (let i = 0; i < n; i++) {
        if (i < k) {
 
            // Printing rightmost
            // kth elements
            document.write(a[n + i - k] + " ");
        }
        else {
 
            // Prints array after
            // 'k' elements
            document.write((a[i - k]) + " ");
        }
    }
    document.write("
");
}
 
// Driver code
let Array = [1, 2, 3, 4, 5];
let N = Array.length;
let K = 2;
 
RightRotate(Array, N, K);
 
// This code is contributed by gfgking.

输出：

4 5 1 2 3

时间复杂度：O(n)
辅助空间：O(1)

如果您想与行业专家一起参加直播课程，请参阅Geeks Classes Live