链表顺时针旋转的Java程序

给定一个单链表和一个整数K ，任务是将链表顺时针向右旋转K位。
例子：

Input: 1 -> 2 -> 3 -> 4 -> 5 -> NULL, K = 2 
Output: 4 -> 5 -> 1 -> 2 -> 3 -> NULL
Input: 7 -> 9 -> 11 -> 13 -> 3 -> 5 -> NULL, K = 12 
Output: 7 -> 9 -> 11 -> 13 -> 3 -> 5 -> NULL 
 

方法：旋转链表首先检查给定的k是否大于链表中的节点数。遍历链表并找到链表的长度，然后将其与 k 进行比较，如果小于则继续，否则通过对链表的长度取模，在链表大小的范围内推导它。
之后从列表的长度中减去 k 的值。现在，问题已更改为链表的左旋转，因此请按照以下步骤操作：

• 将第 k 个节点的下一个节点更改为 NULL。
• 将最后一个节点的下一个节点更改为上一个头节点。
• 将头部更改为第 (k+1) 个节点。

为此，需要指向第 k 个节点、第 (k+1) 个节点和最后一个节点的指针。
下面是上述方法的实现：

Java

// Java implementation of the approach
class GFG
{
     
/* Link list node */
static class Node
{
    int data;
    Node next;
}
 
/* A utility function to push a node */
static Node push(Node head_ref, int new_data)
{
    /* allocate node */
    Node new_node = new Node();
 
    /* put in the data */
    new_node.data = new_data;
 
    /* link the old list off the new node */
    new_node.next = (head_ref);
 
    /* move the head to point to the new node */
    (head_ref) = new_node;
    return head_ref;
}
 
/* A utility function to print linked list */
static void printList(Node node)
{
    while (node != null)
    {
        System.out.print(node.data + " -> ");
        node = node.next;
    }
    System.out.print( "null");
}
 
// Function that rotates the given linked list
// clockwise by k and returns the updated
// head pointer
static Node rightRotate(Node head, int k)
{
 
    // If the linked list is empty
    if (head == null)
        return head;
 
    // len is used to store length of the linked list
    // tmp will point to the last node after this loop
    Node tmp = head;
    int len = 1;
    while (tmp.next != null)
    {
        tmp = tmp.next;
        len++;
    }
 
    // If k is greater than the size
    // of the linked list
    if (k > len)
        k = k % len;
 
    // Subtract from length to convert
    // it into left rotation
    k = len - k;
 
    // If no rotation needed then
    // return the head node   
    if (k == 0 || k == len)
        return head;
 
    // current will either point to
    // kth or null after this loop
    Node current = head;
    int cnt = 1;
    while (cnt < k && current != null)
    {
        current = current.next;
        cnt++;
    }
 
    // If current is null then k is equal to the
    // count of nodes in the list
    // Don't change the list in this case
    if (current == null)
        return head;
 
    // current points to the kth node
    Node kthnode = current;
 
    // Change next of last node to previous head
    tmp.next = head;
 
    // Change head to (k+1)th node
    head = kthnode.next;
 
    // Change next of kth node to null
    kthnode.next = null;
 
    // Return the updated head pointer
    return head;
}
 
// Driver code
public static void main(String args[])
{
 
    /* The constructed linked list is:
    1.2.3.4.5 */
    Node head = null;
    head = push(head, 5);
    head = push(head, 4);
    head = push(head, 3);
    head = push(head, 2);
    head = push(head, 1);
 
    int k = 2;
 
    // Rotate the linked list
    Node updated_head = rightRotate(head, k);
 
    // Print the rotated linked list
    printList(updated_head);
}
}
 
// This code is contributed by Arnub Kundu

输出：

4 -> 5 -> 1 -> 2 -> 3 -> NULL

时间复杂度： O(n)，其中 n 是链表中的节点数。

辅助空间： O(1)

详情请参考链表顺时针旋转的完整文章！