[LeetCode]143 重排链表

题目描述

给定一个单链表 L：L0→L1→…→Ln-1→Ln ，将其重新排列后变为：L0→Ln→L1→Ln-1→L2→Ln-2→…

你不能只是单纯的改变节点内部的值，而是需要实际的进行节点交换。

示例1

给定链表 1->2->3->4, 重新排列为 1->4->2->3.

示例2

给定链表 1->2->3->4->5, 重新排列为 1->5->2->4->3.

思路

  • 找到链表尾端，交换

代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def reorderList(self, head):
        """
        :type head: ListNode
        :rtype: void Do not return anything, modify head in-place instead.
        """
        if head == None or head.next == None:
            return

        front = head
        rear = head.next
        while rear != None and rear.next != None:
            front = front.next
            rear = rear.next.next

        p = front.next
        front.next = None

        cur = None
        while p != None:
            q = p.next
            p.next = cur
            cur = p 
            p = q

        front = head
        while front != None and cur != None:
            tmp = cur.next  
            cur.next = front.next
            front.next = cur
            front = front.next.next
            cur = tmp