题目链接
难度：中等 类型： 链表
给定一个单链表 L：L0→L1→…→Ln-1→Ln ，
将其重新排列后变为： L0→Ln→L1→Ln-1→L2→Ln-2→…
你不能只是单纯的改变节点内部的值，而是需要实际的进行节点交换。
示例1
给定链表 1->2->3->4, 重新排列为 1->4->2->3.
示例2
给定链表 1->2->3->4->5, 重新排列为 1->5->2->4->3.
解题思路
先找到链表的中点：快慢指针，考虑节点个数奇偶情况
再翻转后半段链表
最后两链表依次插入
代码实现
# 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.
"""
# looking for the midpoint
fast = slow = a = head
while fast and fast.next:
fast = fast.next.next
slow = slow.next
b = slow.next if fast else slow
# reverse
pre = None
cur = b
while cur:
next = cur.next
cur.next = pre
pre = cur
cur = next
# cross
b = pre
while a:
a.next, b = b, a.next
a = a.next
return head
