[LeetCode]题解（python）：143-Reorder List

题目来源：

　　https://leetcode.com/problems/reorder-list/

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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。

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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。

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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