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