Java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。

Java

// Java program to rotate a linked list 
// block wise 
import java.util.*;
class GFG{
  
// Link list node 
static class Node 
{ 
    int data; 
    Node next; 
}; 
static Node tail;
    
// Recursive function to rotate one block 
static Node rotateHelper(Node blockHead, 
                         Node blockTail, 
                         int d, int k) 
{ 
    if (d == 0) 
        return blockHead; 
  
    // Rotate Clockwise 
    if (d > 0) 
    { 
        Node temp = blockHead; 
        for (int i = 1; temp.next.next!=null && 
                 i < k - 1; i++) 
            temp = temp.next; 
        blockTail.next = blockHead; 
        tail = temp; 
        return rotateHelper(blockTail, temp, 
                            d - 1,  k); 
    } 
  
    // Rotate anti-Clockwise 
    if (d < 0) 
    { 
        blockTail.next = blockHead; 
        tail = blockHead; 
        return rotateHelper(blockHead.next, 
                            blockHead, d + 1,  k); 
    } 
    return blockHead;
} 
  
// Function to rotate the linked list 
// block-wise 
static Node rotateByBlocks(Node head, 
                           int k, int d) 
{ 
    // If length is 0 or 1 return head 
    if (head == null || head.next == null) 
        return head; 
  
    // if degree of rotation is 0, 
    // return head 
    if (d == 0) 
        return head; 
  
    Node temp = head;
    tail = null; 
  
    // Traverse upto last element of 
    // this block 
    int i; 
    for (i = 1; temp.next != null &&
         i < k; i++) 
        temp = temp.next; 
  
    // Storing the first node of next block 
    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, i); 
    else
        head = rotateHelper(head, temp, 
                            d % k, 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 
static Node push(Node head_ref, 
                 int new_data) 
{ 
    Node new_node = new Node(); 
    new_node.data = new_data; 
    new_node.next = head_ref; 
    head_ref = new_node;
    return head_ref;
      
} 
  
// Function to print linked list 
static void printList(Node node) 
{ 
    while (node != null) 
    { 
        System.out.print(node.data + " "); 
        node = node.next; 
    } 
} 
  
// Driver code
public static void main(String[] args) 
{ 
    
    /* Start with the empty list */
    Node head = null; 
  
    // Create a list 1.2.3.4.5. 
    // 6.7.8.9.null 
    for (int i = 9; i > 0; i -= 1) 
        head = push(head, i); 
    System.out.print("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); 
    System.out.print(
    "Rotated by blocks Linked list "); 
    printList(head); 
}
} 
// This code is contributed by aashish1995

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！