Leetcode 61: 旋转链表

题目描述

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 * $10^9$

思路

由于k的值可能很大，所以要先将k除len取余。
本题的实质是找到倒数第k个结点，将其后面所有结点调到链表前面。
找到倒数第k个结点：定义两个结点，让其相隔k个结点，后一起往后移动。

代码

class Solution:
    def rotateRight(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:
        l = 0
        cur = head
        while cur is not None:
            l+=1
            cur = cur.next
        k = k%l
        bef = head
        aft = head
        for i in range(k):
            aft = aft.next
        while(aft.next is not None):
            aft = aft.next
            bef = bef.next

        # print(bef.val, aft.val)

        aft.next = head
        res = bef.next
        bef.next = None
        return res