用于数组旋转的块交换算法的 C# 程序

原文:https://www . geesforgeks . org/csharp-程序换块-交换-算法换阵列-旋转/

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

Array

将上面的数组旋转 2 将构成数组

ArrayRotation1

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: 这里是同样算法的迭代实现。这里使用了相同的实用函数 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) 其他阵轮换法请看以下帖子: https://www.geeksforgeeks.org/array-rotation/ https://www . geeksforgeeks . org/程序换阵-轮换-续-反转-算法/

参考文献: 【http://www.cs.bell-labs.com/cm/cs/pearls/s02b.pdf】 如发现以上程序/算法有任何 bug 或想分享任何关于区块交换算法的补充信息,请写评论。

更多详情请参考数组旋转的块交换算法的完整文章!