1 class Solution: 2 def reorderList(self, head: ListNode) -> None: 3 """ 4 Do not return anything, modify head in-place instead. 5 """ 6 if not head: 7 return 8 aux = self.findmid(head) 9 new_head = self.reverse(aux.next) 10 aux.next = None 11 l, r = head, new_head 12 while l and r: 13 l_next = l.next 14 r_next = r.next 15 l.next = r 16 r.next = l_next 17 l = l_next 18 r = r_next 19 20 def findmid(self, head): 21 if not head: 22 return 23 fast = slow = head 24 while fast and fast.next: 25 fast = fast.next.next 26 slow = slow.next 27 return slow 28 def reverse(self, head): 29 if not head: 30 return None 31 pre = None 32 while head: 33 tmp = head.next 34 head.next = pre 35 pre = head 36 head = tmp 37 return pre
先找到链表的中间节点，然后从中间节点将链表一分为二。
将后半部分的节点倒序。然后轮流从两个链表中选择节点连接在一起。
参考：https://leetcode.com/problems/reorder-list/discuss/447242/python-solution