快速查找数组的多个左旋转 |设置 1

给定一个大小为 n 的数组和多个值，我们需要围绕该数组左旋。如何快速找到多个左旋转？
例子：

Input : arr[] = {1, 3, 5, 7, 9}
        k1 = 1
        k2 = 3
        k3 = 4
        k4 = 6
Output : 3 5 7 9 1
         7 9 1 3 5
         9 1 3 5 7
         3 5 7 9 1

Input : arr[] = {1, 3, 5, 7, 9}
        k1 = 14 
Output : 9 1 3 5 7

简单方法：我们已经讨论了以下帖子中给出的不同方法。

1. 数组的左旋转（简单和杂耍算法）。
2. 数组旋转的块交换算法
3. 数组旋转的反转算法

上述方法中最好的方法需要 O(n) 时间和 O(1) 额外空间。
有效的方法：
当需要单次旋转时，上述方法效果很好。这些方法还修改了原始数组。为了处理数组旋转的多个查询，我们使用大小为 2n 的临时数组并快速处理旋转。
第 1 步：在 temp[0..2n-1] 数组中复制整个数组两次。
步骤 2：在 temp[] 中经过 k 次旋转后数组的起始位置将是 k % n。我们做
第 3 步：将 temp[] 数组从 k % n 打印到 k % n + n。

C++

// CPP implementation of left rotation of
// an array K number of times
#include
using namespace std;
 
// Fills temp[] with two copies of arr[]
void preprocess(int arr[], int n, int temp[])
{
    // Store arr[] elements at i and i + n
    for (int i = 0; i

---

Java

// Java implementation of left rotation of
// an array K number of times
class LeftRotate
{
    // Fills temp[] with two copies of arr[]
    static void preprocess(int arr[], int n,
                                   int temp[])
    {
        // Store arr[] elements at i and i + n
        for (int i = 0; i

---

Python3

# Python3 implementation of left rotation
# of an array K number of times
 
# Fills temp with two copies of arr
def preprocess(arr, n):
    temp = [None] * (2 * n)
     
    # Store arr elements at i and i + n
    for i in range(n):
        temp[i] = temp[i + n] = arr[i]
    return temp
 
# Function to left rotate an array k times
def leftRotate(arr, n, k, temp):
     
    # Starting position of array after k
    # rotations in temp will be k % n
    start = k % n
     
    # Print array after k rotations
    for i in range(start, start + n):
        print(temp[i], end = " ")
    print("")
 
# Driver program
arr = [1, 3, 5, 7, 9]
n = len(arr)
temp = preprocess(arr, n)
 
k = 2
leftRotate(arr, n, k, temp)
       
k = 3
leftRotate(arr, n, k, temp)
       
k = 4
leftRotate(arr, n, k, temp)
 
# This code is contributed by Sanghamitra Mishra

C#

// C# implementation of left rotation of
// an array K number of times
using System;
class LeftRotate
{
    // Fills temp[] with two copies of arr[]
    static void preprocess(int []arr, int n,
                                int[] temp)
    {
        // Store arr[] elements at i and i + n
        for (int i = 0; i

---

Javascript

 
// Javascript implementation of left rotation of
// an array K number of times
 
    // Fills temp with two copies of arr
    function preprocess(arr , n , temp) {
        // Store arr elements at i and i + n
        for (i = 0; i < n; i++)
            temp[i] = temp[i + n] = arr[i];
    }
 
    // Function to left rotate an array k time
    function leftRotate(arr , n , k , temp) {
        // Starting position of array after k
        // rotations in temp will be k % n
        var start = k % n;
 
        // Print array after k rotations
        for (i = start; i < start + n; i++)
            document.write(temp[i] + " ");
 
        document.write("<br/>");
    }
 
    // Driver program
     
        var arr = [ 1, 3, 5, 7, 9 ];
        var n = arr.length;
 
        var temp = Array(2 * n).fill(0);
        preprocess(arr, n, temp);
 
        var k = 2;
        leftRotate(arr, n, k, temp);
 
        k = 3;
        leftRotate(arr, n, k, temp);
 
        k = 4;
        leftRotate(arr, n, k, temp);
 
// This code contributed by gauravrajput1
 

输出：

5 7 9 1 3 
7 9 1 3 5 
9 1 3 5 7 

请注意，查找旋转起始地址的任务需要O(1) 时间。它正在打印需要 O(n) 时间的元素。
空间优化方法：上述方法需要额外的空间。下面给出的是一个空间优化的解决方案。感谢frenzy77提出这种方法。


C++

// CPP implementation of left rotation of
// an array K number of times
#include
using namespace std;
 
// Function to left rotate an array k times
void leftRotate(int arr[], int n, int k)
{
    // Print array after k rotations
    for (int i = k; i < k + n; i++)
        cout << arr[i%n] << " ";
}
 
// Driver program
int main()
{
    int arr[] = {1, 3, 5, 7, 9};
    int n = sizeof(arr) / sizeof(arr[0]);
 
    int k = 2;
    leftRotate(arr, n, k);
    cout << endl;
 
    k = 3;
    leftRotate(arr, n, k);
    cout << endl;
 
    k = 4;
    leftRotate(arr, n, k);
    cout << endl;
 
    return 0;
}

Java

// Java implementation of
// left rotation of an
// array K number of times
 
import java.io.*;
 
class GFG
{
 
// Function to left rotate
// an array k times
static void leftRotate(int arr[],
                       int n, int k)
{
    // Print array after
    // k rotations
    for (int i = k; i < k + n; i++)
        System.out.print(arr[i % n] + " ");
}
 
// Driver Code
public static void main (String[] args)
{
    int arr[] = {1, 3, 5, 7, 9};
    int n = arr.length;
     
    int k = 2;
    leftRotate(arr, n, k);
    System.out.println();
     
    k = 3;
    leftRotate(arr, n, k);
    System.out.println();
     
    k = 4;
    leftRotate(arr, n, k);
    System.out.println();
}
}
 
// This code is contributed by ajit

Python3

# Python3 implementation of
# left rotation of an array
# K number of times
 
# Function to left rotate
# an array k times
def leftRotate(arr, n, k):
     
    # Print array
    # after k rotations
    for i in range(k, k + n):
        print(str(arr[i % n]),
                   end = " ")
 
# Driver Code
arr = [1, 3, 5, 7, 9]
n = len(arr)
k = 2;
leftRotate(arr, n, k)
print()
 
k = 3;
leftRotate(arr, n, k)
print()
 
k = 4
leftRotate(arr, n, k)
print()
 
# This code is contributed
# by ChitraNayal

C＃

// C# implementation of
// left rotation of an
// array K number of times
using System;
 
class GFG
{
 
// Function to left rotate
// an array k times
static void leftRotate(int []arr,
                       int n, int k)
{
    // Print array after
    // k rotations
    for (int i = k; i < k + n; i++)
        Console.Write(arr[i % n] + " ");
}
 
// Driver Code
static public void Main ()
{
int []arr = {1, 3, 5, 7, 9};
int n = arr.Length;
 
int k = 2;
leftRotate(arr, n, k);
Console.WriteLine();
 
k = 3;
leftRotate(arr, n, k);
Console.WriteLine();
 
k = 4;
leftRotate(arr, n, k);
Console.WriteLine();
}
}
 
// This code is contributed
// by akt_mit

Javascript

 
// JavaScript implementation of
// left rotation of an
// array K number of times
 
 
// Function to left rotate
// an array k times
function leftRotate(arr, n, k)
{
    // Print rray after
    // k rotations
    for (let i = k; i < k + n; i++)
        document.write(arr[i % n] + " ");
}
 
// Driver Code
 
let arr = [1, 3, 5, 7, 9];
n = arr.length;
 
k = 2;
leftRotate(arr, n, k);
document.write("<br>");
 
k = 3;
leftRotate(arr, n, k);
document.write("<br>");
 
k = 4;
leftRotate(arr, n, k);
document.write("<br>");
 
 
 

输出：

5 7 9 1 3 
7 9 1 3 5 
9 1 3 5 7