Leetcode 中等：61. 旋转链表

题目：旋转链表

• 题号：61
• 难度：中等
• https://leetcode-cn.com/problems/rotate-list/

给你一个链表的头节点 head ，旋转链表，将链表每个节点向右移动 k 个位置。

示例 1:

输入：head = [1,2,3,4,5], k = 2
输出：[4,5,1,2,3]

示例 2:

输入：head = [0,1,2], k = 4
输出：[2,0,1]

提示：

• 链表中节点的数目在范围 [0, 500] 内
• -100 <= Node.val <= 100
• $0<=k<=2\ast 10^9$

---

实现

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int x) { val = x; }
 * }
 */
 
public class Solution
{
    public ListNode RotateRight(ListNode head, int k)
    {
        if (head == null || k == 0)
            return head;

        int len = GetLength(head);
        int index = len - k%len;

        if (index == len)
            return head;

        ListNode temp1 = head;
        ListNode temp2 = head;
        for (int i = 0; i < index - 1; i++)
        {
            temp1 = temp1.next;
        }
        head = temp1.next;
        temp1.next = null;

        temp1 = head;
        while (temp1.next != null)
        {
            temp1 = temp1.next;
        }
        temp1.next = temp2;
        return head;
    }

    public int GetLength(ListNode head)
    {
        ListNode temp = head;
        int i = 0;
        while (temp != null)
        {
            i++;
            temp = temp.next;
        }
        return i;
    }
} 

Python 语言

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution(object):
    def rotateRight(self, head, k):
        """
        :type head: ListNode
        :type k: int
        :rtype: ListNode
        """
        if head is None or k == 0:
            return head

        l = self.GetLength(head)
        index = l - k % l
        if index == l:
            return head

        temp1 = temp2 = head
        for i in range(index - 1):
            temp1 = temp1.next
        head = temp1.next
        temp1.next = None

        temp1 = head
        while temp1.next is not None:
            temp1 = temp1.next
        temp1.next = temp2
        return head

    def GetLength(self, head):
        temp = head
        i = 0
        while temp is not None:
            i += 1
            temp = temp.next
        return i