C程序以块方式旋转链表

给定一个长度为 n 和块长度为k的链表，每个块以循环方式向右/向左旋转一个数字d 。如果d为正，则向右旋转，否则向左旋转。

例子：

Input: 1->2->3->4->5->6->7->8->9->NULL, 
        k = 3 
        d = 1
Output: 3->1->2->6->4->5->9->7->8->NULL
Explanation: Here blocks of size 3 are
rotated towards right(as d is positive) 
by 1.
 
Input: 1->2->3->4->5->6->7->8->9->10->
               11->12->13->14->15->NULL, 
        k = 4 
        d = -1
Output: 2->3->4->1->6->7->8->5->10->11
             ->12->9->14->15->13->NULL
Explanation: Here, at the end of linked 
list, remaining nodes are less than k, i.e.
only three nodes are left while k is 4. 
Rotate those 3 nodes also by d.

先决条件：旋转链表
这个想法是如果 d 的绝对值大于 k 的值，则将链表旋转 d % k 次。如果 d 为 0，则根本不需要旋转链表。

C

// C program to rotate a linked list 
// block wise
#include 
#include 
  
// Link list node 
struct Node 
{
    int data;
    struct Node* next;
};
  
// Recursive function to rotate one block
struct Node* rotateHelper(struct Node* blockHead,
                          struct Node* blockTail,
                          int d, struct Node** tail,
                          int k)
{
    if (d == 0)
        return blockHead;
  
    // Rotate Clockwise
    if (d > 0) 
    {
        struct Node* temp = blockHead;
        for (int i = 1; temp->next->next &&
                 i < k - 1; i++)
            temp = temp->next;
        blockTail->next = blockHead;
        *tail = temp;
        return rotateHelper(blockTail, temp,
                            d - 1, tail, k);
    }
  
    // Rotate anti-Clockwise
    if (d < 0) 
    {
        blockTail->next = blockHead;
        *tail = blockHead;
        return rotateHelper(blockHead->next,
                            blockHead, d + 1, 
                            tail, k);
    }
}
  
// Function to rotate the linked list 
// block wise
struct Node* rotateByBlocks(struct Node* head,
                            int k, int d)
{
    // If length is 0 or 1 return head
    if (!head || !head->next)
        return head;
  
    // if degree of rotation is 0, 
    // return head
    if (d == 0)
        return head;
  
    struct Node* temp = head, 
               *tail = NULL;
  
    // Traverse upto last element of 
    // this block
    int i;
    for (i = 1; temp->next && i < k; i++)
        temp = temp->next;
  
    // Storing the first node of 
    // next block
    struct Node* nextBlock = temp->next;
  
    // If nodes of this block are less 
    // than k.
    // Rotate this block also
    if (i < k)
        head = rotateHelper(head, temp, 
                            d % k, &tail, i);
    else
        head = rotateHelper(head, temp, 
                            d % k, &tail, k);
  
    // Append the new head of next block
    // to the tail of this block
    tail->next = rotateByBlocks(nextBlock, k,
                                d % k);
  
    // return head of updated Linked List
    return head;
}
  
// UTILITY FUNCTIONS 
// Function to push a node 
void push(struct Node** head_ref, 
          int new_data)
{
    struct Node* new_node = new Node;
    new_node->data = new_data;
    new_node->next = (*head_ref);
    (*head_ref) = new_node;
}
  
// Function to print linked list 
void printList(struct Node* node)
{
    while (node != NULL) 
{
        printf("%d ", node->data);
        node = node->next;
    }
}
  
// Driver code
int main()
{
    // Start with the empty list 
    struct Node* head = NULL;
  
    // Create a list 1->2->3->4->5->
    // 6->7->8->9->NULL
    for (int i = 9; i > 0; i -= 1)
        push(&head, i);
  
    printf("Given linked list ");
    printList(head);
  
    // k is block size and d is number of
    // rotations in every block.
    int k = 3, d = 2;
    head = rotateByBlocks(head, k, d);
  
    printf(
    "Rotated by blocks Linked list ");
    printList(head);
  
    return (0);
}

输出：

Given linked list 
1 2 3 4 5 6 7 8 9 
Rotated by blocks Linked list 
2 3 1 5 6 4 8 9 7

有关详细信息，请参阅有关 Rotate Linked List 块的完整文章！