右旋转 K 次后打印数组|设置 2
原文:https://www . geesforgeks . org/print-array-it-right-rotated-k-times-set-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。
示例:
输入: arr = {1，3，5，7，9}，K = 2 输出: 7 9 1 3 5
输入: arr = {1，2，3，4，5}，K = 4 T3】输出: 2 3 4 5 1
算法:给定的问题可以通过反转子阵来解决。可以遵循以下步骤来解决问题:
- 将所有数组元素从 1 反转到 N -1
- 将数组元素从 1 反转到K–1
- 将数组元素从 K 反转到 N -1
C++
// C++ implementation for the above approach
#include <bits/stdc++.h>
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 语言(一种计算机语言，尤用于创建网站)
/*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.
Python 3
# 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.
java 描述语言
<script>
// Javascript program for the above approach
// Function to rotate the array
// to the right, K times
function RightRotate(Array, N, K)
{
// Reverse all the array elements
for (let i = 0; i < N / 2; i++) {
let temp = Array[i];
Array[i] = Array[N - i - 1];
Array[N - i - 1] = temp;
}
// Reverse the first k elements
for (let i = 0; i < K / 2; i++) {
let 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 (let i = 0; i < (K + N) / 2; i++) {
let 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 (let i = 0; i < N; i++) {
document.write(Array[i] + " ");
}
}
// Driver Code
let arr = [ 1, 2, 3, 4, 5 ];
let N = arr.length;
let K =4;
// Call the function and
// print the answer
RightRotate(arr, N, K);
// This code is contributed by Samim Hossain Mondal.
</script>
Output
2 3 4 5 1
时间复杂度 : O(N) 辅助空间 : O(1)
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