用于数组旋转的块交换算法的 C# 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：

C#

using System;
  
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);     
    } 
      
    // If B is shorter
    else 
    { 
        swap(arr, i, d, n - d); 
          
        // This is tricky
        leftRotateRec(arr, n - d + i, 
                       2 * d - n, d); 
    } 
}
  
// UTILITY FUNCTIONS
// Function to print an array 
public static void printArray(int []arr,
                              int size) 
{ 
    int i; 
    for(i = 0; i < size; i++) 
        Console.Write(arr[i] + " "); 
          
    Console.WriteLine(); 
} 
  
// 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 amal kumar choubey

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()。

C＃

// C# 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;
        }
          
    }
      
    /*Finally, block swap A and B*/
    swap(arr, d-i, d, i);
}
  
// This code is contributed by Rajput-Ji

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！