题目如下：
Given a singly linked list L: L0→L1→…→Ln-1→Ln,
reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…You may not modify the values in the list‘s nodes, only nodes itself may be changed.
Example 1:
Given 1->2->3->4, reorder it to 1->4->2->3.Example 2:
Given 1->2->3->4->5, reorder it to 1->5->2->4->3.
解题思路：我的方法是用一个list保存链表后半部分的节点，接下来从头开始遍历链表，把list中的第一个节点插入到原链表的第一个节点后，第二个插入到原链表的第二个节点后面，直到list全部节点都插入到原链表为止。
代码如下：
# 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: None Do not return anything, modify head in-place instead. """ if head == None: return mid,tail = head,head while tail != None: tail = tail.next if tail == None: break tail = tail.next if tail == None: break mid = mid.next l = [] second_half = mid.next mid.next = None while second_half != None: l.insert(0,second_half) second_half = second_half.next current = head while len(l) > 0: node = l.pop(0) node.next = current.next current.next = node current = current.next.next
原文：https://www.cnblogs.com/seyjs/p/10559919.html