【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 = None

class 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