题目：

题解：

class Solution:# 翻转一个子链表，并且返回新的头与尾def reverse(self, head: ListNode, tail: ListNode):prev = tail.nextp = headwhile prev != tail:nex = p.nextp.next = prevprev = pp = nexreturn tail, headdef reverseKGroup(self, head: ListNode, k: int) -> ListNode:hair = ListNode(0)hair.next = headpre = hairwhile head:tail = pre# 查看剩余部分长度是否大于等于 kfor i in range(k):tail = tail.nextif not tail:return hair.nextnex = tail.nexthead, tail = self.reverse(head, tail)# 把子链表重新接回原链表pre.next = headtail.next = nexpre = tailhead = tail.nextreturn hair.next