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）：
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None class Solution(object):
def reorderList(self, head):
"""
:type head: ListNode
:rtype: void Do not return anything, modify head in-place instead.
"""
if head == None or head.next == None or head.next.next == None:
return
slow = fast = head
while fast and fast.next:
slow = slow.next
fast = fast.next.next
head1,head2 = head,slow.next
slow.next = None tmp = ListNode(0)
tmp.next,p = head2,head2.next
head2.next = None
while p:
tmp1,p = p,p.next
tmp1.next = tmp.next
tmp.next = tmp1
head2 = tmp.next while head2:
tmp2,tmp3 = head1.next,head2.next
head1.next = head2
head2.next = tmp2
head1,head2 = tmp2,tmp3