leetcode 【 Reorder List 】python 实现

题目:

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}.

代码: oj 测试通过 248 ms

 # 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 is None or head.next is None or head.next.next is None:
             return head

         dummyhead = ListNode(0)
         dummyhead.next = head

         # get the length of the linked list
         p = head
         list_length = 0
         while p is not None:
             list_length += 1
             p = p.next

         #reverse the second half linked list
         fast = dummyhead
         for i in range((list_length+1)/2):
             fast = fast.next
         pre = fast
         curr = pre.next
         for i in range( (list_length)/2 - 1 ):
             tmp = curr.next
             curr.next = tmp.next
             tmp.next = pre.next
             pre.next = tmp

         #merge
         h2 = pre.next
         fast.next = None # cut the connection between 1st half linked list and 2nd half linked list
         while head is not None and h2 is not None:
             tmp = head.next
             head.next = h2
             tmp2 = h2.next
             head.next.next = tmp
             h2 = tmp2
             head = tmp

         return dummyhead.next

思路：

这道题的路子分三块：

1. 遍历单链表 求链表长度

2. 锁定后半个链表，反转后半个链表的每个元素

3. 切断前后半个链表的链接处 然后合并两个链表