61. 旋转链表

Difficulty
Medium
Tags
链表
URL
https://leetcode.cn/problems/rotate-list/
Star
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：
notion imagenotion image
输入：head = [1,2,3,4,5], k = 2
输出：[4,5,1,2,3]
示例 2：
notion imagenotion image
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 * 109
通过次数247,758提交次数594,352

法1

思路
计算倒数 k 个节点 转化为计算正数 length-k 个节点，然后找到之后，接到头节点
题解
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def rotateRight(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:
        if head is None:
            return head
        length = 0
        cur = head
        while cur:
            cur = cur.next
            length += 1
        
        k = k % length
        # k = 0 表示不用翻转
        if k == 0:
            return head

        # 找倒数第二个节点 = 找正数 第 length - k 个 节点
        k = length - k
        count = 1
        cur = head
        while cur:
            if count == k:
                break
            count += 1
            cur = cur.next
        # 把后半段保存一下，再断开
        nxt = cur.next
        cur.next = None
        
        res = nxt
        # 找到后半段的尾巴，接到 head 上
        while nxt and nxt.next:
            nxt = nxt.next
        nxt.next = head

        return res