右旋转 K 次后打印数组
原文:https://www . geesforgeks . org/print-array-it-right-rotated-k-times/
给定一个大小为 N 的数组和一个值为 K ，我们需要围绕它右旋转数组。如何快速打印右旋转数组？ 例:
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
进场:
-
我们将首先取 K 乘以 N (K = K % N)的 mod，因为在每 N 次旋转之后，数组将变得与初始数组相同。
-
现在，我们将数组从 i = 0 迭代到 i = N-1 并检查，
-
如果 i < K ，打印最右边的第 Kth 个元素(a[N + i -K])。否则，
-
在“K”元素后打印数组(a[I–K])。
-
下面是上述方法的实现。
C++
// C++ implementation of right rotation
// of an array K number of times
#include<bits/stdc++.h>
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 语言(一种计算机语言，尤用于创建网站)
// 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<k)
{
// Printing rightmost
// kth elements
System.out.print(a[n + i - k]
+ " ");
}
else
{
// Prints array after
// 'k' elements
System.out.print(a[i - k]
+ " ");
}
}
System.out.println();
}
// Driver program
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 M Vamshi Krishna
Python 3
# 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
java 描述语言
// 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("<br>");
}
// 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.
Output:
4 5 1 2 3
时间复杂度:O(n) T3】辅助空间 : O(1)
请参阅以下帖子了解阵列旋转的其他方法:
https://www . geesforgeks . org/print-array-it-right-rotated-k-times-set-2/
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