【LEETCODE】143- Reorder List [Python]

Given a singly linked list L: L0→L1→…→Ln-1→Ln,

reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…

You must do this in-place without altering the nodes' values.

For example,

Given {1,2,3,4}, reorder it to{1,4,2,3}.

题意：

给一个单链：L: L0→L1→…→Ln-1→Ln,

将它变成：L0→Ln→L1→Ln-1→L2→Ln-2→…

注意in-place，不要改变值，不需要返回值

思路：

在中间平均截开两段，若为奇数，则第一段多一个

将第二段reverse

然后两段再交错地link起来

Python:

# Definition for singly-linked list.# class ListNode(object):#     def __init__(self, x):#         self.val = x#         self.next = Noneclass Solution(object):    def reorderList(self, head):        """        :type head: ListNode        :rtype: void Do not return anything, modify head in-place instead.        """        if head is None or head.next is None or head.next.next is None:            head=head        else:                    slow=fast=head                                      # two parts            while fast.next and fast.next.next:                slow=slow.next                fast=fast.next.next                    head2=slow.next            slow.next=None                    dummy=ListNode(0)                                   # reverse 2nd part            dummy.next=head2            p=head2.next            head2.next=None            while p:                tmp=p                p=p.next                tmp.next=dummy.next                dummy.next=tmp            head2=dummy.next                    p1=head                                             # rejoin 2 parts together            p2=head2            while p2:                t1=p1.next                p1.next=p2                t2=p2.next                p2.next=t1                p1=t1                p2=t2