用于数组旋转的块交换算法的Java程序

编写一个函数rotate(ar[], d, n) 将大小为 n 的 arr[] 旋转 d 个元素。

将上述数组旋转 2 次将生成数组

算法 ：

Initialize A = arr[0..d-1] and B = arr[d..n-1]
1) Do following until size of A is equal to size of B

  a)  If A is shorter, divide B into Bl and Br such that Br is of same 
       length as A. Swap A and Br to change ABlBr into BrBlA. Now A
       is at its final place, so recur on pieces of B.  

   b)  If A is longer, divide A into Al and Ar such that Al is of same 
       length as B Swap Al and B to change AlArB into BArAl. Now B
       is at its final place, so recur on pieces of A.

2)  Finally when A and B are of equal size, block swap them.

递归实现：

Java

import java.util.*;
  
class GFG 
{
    // Wrapper over the recursive function leftRotateRec()
    // It left rotates arr[] by d.
    public static void leftRotate(int arr[], int d, 
                                                int n)
    {
        leftRotateRec(arr, 0, d, n);
    }
  
    public static void leftRotateRec(int arr[], int i, 
                                  int d, int n)
    {
        /* Return If number of elements to be rotated 
        is zero or equal to array size */
        if(d == 0 || d == n) 
            return; 
          
        /*If number of elements to be rotated 
        is exactly half of array size */
        if(n - d == d) 
        { 
            swap(arr, i, n - d + i, d); 
            return; 
        } 
          
        /* If A is shorter*/    
        if(d < n - d) 
        { 
            swap(arr, i, n - d + i, d); 
            leftRotateRec(arr, i, d, n - d);     
        } 
        else /* If B is shorter*/    
        { 
            swap(arr, i, d, n - d); 
            leftRotateRec(arr, n - d + i, 2 * d - n, d); /*This is tricky*/
        } 
    }
  
/*UTILITY FUNCTIONS*/
/* function to print an array */
public static void printArray(int arr[], int size) 
{ 
    int i; 
    for(i = 0; i < size; i++) 
        System.out.print(arr[i] + " "); 
    System.out.println(); 
} 
  
/*This function swaps d elements 
starting at index fi with d elements
starting at index si */
public static void swap(int arr[], int fi,
                        int si, int d) 
{ 
    int i, temp; 
    for(i = 0; i < d; i++) 
    { 
        temp = arr[fi + i]; 
        arr[fi + i] = arr[si + i]; 
        arr[si + i] = temp; 
    } 
} 
  
// Driver Code
public static void main (String[] args) 
{
    int arr[] = {1, 2, 3, 4, 5, 6, 7}; 
    leftRotate(arr, 2, 7); 
    printArray(arr, 7);     
}
}
  
// This code is contributed by codeseeker

输出：

3 5 4 6 7 1 2

迭代实现：
这是相同算法的迭代实现。此处使用相同的实用函数swap()。

Java

//Java code for above implementation
static void leftRotate(int arr[], int d, int n) 
{ 
int i, j; 
if(d == 0 || d == n) 
    return; 
i = d; 
j = n - d; 
while (i != j) 
{ 
    if(i < j) /*A is shorter*/
    { 
    swap(arr, d-i, d+j-i, i); 
    j -= i; 
    } 
    else /*B is shorter*/
    { 
    swap(arr, d-i, d, j); 
    i -= j; 
    } 
    // printArray(arr, 7); 
} 
/*Finally, block swap A and B*/
swap(arr, d-i, d, i); 
} 

时间复杂度： O(n)
有关数组旋转的其他方法，请参阅以下帖子：
https://www.geeksforgeeks.org/array-rotation/
https://www.geeksforgeeks.org/program-for-array-rotation-continued-reversal-algorithm/

有关更多详细信息，请参阅有关数组旋转的块交换算法的完整文章！