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原题地址：http://oj.leetcode.com/problems/reorder-list/
题意：
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}.
解题思路：1，先将链表截断为两个相等长度的链表，如果链表长度为奇数，则第一条链表长度多1。如原链表为L={1,2,3,4,5}，那么拆分结果为L1={1,2,3}；L2={4,5}。拆分的技巧还是快慢指针的技巧。
2，将第二条链表L2翻转，如将L2={4,5}翻转为L2={5,4}。
3，按照题意归并链表。
# Definition for singly-linked list. # class ListNode: # def __init__(self, x): # self.val = x # self.next = None class Solution: # @param head, a ListNode # @return nothing def reorderList(self, head): if head==None or head.next==None or head.next.next==None: return head # break linked list into two equal length slow = fast = head #快慢指针技巧 while fast and fast.next: #需要熟练掌握 slow = slow.next #链表操作中常用 fast = fast.next.next head1 = head head2 = slow.next slow.next = None # reverse linked list head2 dummy=ListNode(0); dummy.next=head2 #翻转前加一个头结点 p=head2.next; head2.next=None #将p指向的节点一个一个插入到dummy后面 while p: #就完成了链表的翻转 tmp=p; p=p.next #运行时注意去掉中文注释 tmp.next=dummy.next dummy.next=tmp head2=dummy.next # merge two linked list head1 and head2 p1 = head1; p2 = head2 while p2: tmp1 = p1.next; tmp2 = p2.next p1.next = p2; p2.next = tmp1 p1 = tmp1; p2 = tmp2
[leetcode]Reorder List @ Python,布布扣,bubuko.com
[leetcode]Reorder List @ Python
标签：des style blog class code java ext javascript color int rgb
原文地址：http://www.cnblogs.com/zuoyuan/p/3700846.html
