题目来源：
https://leetcode.com/problems/reorder-list/
题意分析：
给定一个链表L：L0→L1→…→Ln-1→Ln，改变链表的排序为： L0→Ln→L1→Ln-1→L2→Ln-2→…，要求时间复杂度为O（n），不能改变节点的值。
题目思路：
题目思路是把链表拆成两个长度相等的链表，然后将后面的链表翻转，重新连起来。
代码（python）：
1 # Definition for singly-linked list. 2 # class ListNode(object): 3 # def __init__(self, x): 4 # self.val = x 5 # self.next = None 6 7 class Solution(object): 8 def reorderList(self, head): 9 """ 10 :type head: ListNode 11 :rtype: void Do not return anything, modify head in-place instead. 12 """ 13 if head == None or head.next == None or head.next.next == None: 14 return 15 slow = fast = head 16 while fast and fast.next: 17 slow = slow.next 18 fast = fast.next.next 19 head1,head2 = head,slow.next 20 slow.next = None 21 22 tmp = ListNode(0) 23 tmp.next,p = head2,head2.next 24 head2.next = None 25 while p: 26 tmp1,p = p,p.next 27 tmp1.next = tmp.next 28 tmp.next = tmp1 29 head2 = tmp.next 30 31 while head2: 32 tmp2,tmp3 = head1.next,head2.next 33 head1.next = head2 34 head2.next = tmp2 35 head1,head2 = tmp2,tmp3