【LeetCode】旋转链表

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一、题目

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

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二、解法

先取余，化简一下k，然后找到倒数第k+1个节点，然后做断开重新连接的动作

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完整代码

# 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 not head:return headdummy = cur = ListNode()dummy.next = headtot = 0while cur.next:cur = cur.nexttot += 1last = curcur = dummyk %= tottmp = curfor _ in range(k + 1):tmp = tmp.nextkk = curwhile tmp:kk = kk.nexttmp = tmp.nextnewHead = kk.nextif not newHead:return dummy.nextlast.next = headkk.next = Nonereturn newHead

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