LeetCode 61 旋转链表

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LeetCode 61 旋转链表

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 * 109

解题思路：

闭合为环

力扣

【Leetcode】Python&C++：61. 旋转链表 (链表操作)【每日一题系列20210327】_哔哩哔哩_bilibili

Python代码：

# 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 k == 0 or not head or not head.next:return headcur = headn = 1while cur.next:cur = cur.nextn += 1add = n - k % nif add == n:return headcur.next = headwhile add:cur = cur.nextadd -= 1res = cur.nextcur.next = Nonereturn res