leetcode 【 Remove Duplicates from Sorted List II 】 python 实现

最新推荐文章于 2025-09-13 23:06:41 发布

转载最新推荐文章于 2025-09-13 23:06:41 发布 · 62 阅读

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原文链接：http://www.cnblogs.com/xbf9xbf/p/4186852.html

文章标签：

#python #数据结构与算法

本文提供了一种解决方案，用于从已排序的链表中移除所有具有重复数值的节点，仅保留原始列表中的唯一数值。该方法通过设置虚拟头结点简化了链表操作，并采用双循环结构来跳过所有重复元素。

题目：

Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.

For example,
Given 1->2->3->3->4->4->5, return 1->2->5.
Given 1->1->1->2->3, return 2->3.

代码：oj在线测试通过 288 ms

 1 # Definition for singly-linked list.
 2 # class ListNode:
 3 #     def __init__(self, x):
 4 #         self.val = x
 5 #         self.next = None
 6 
 7 class Solution:
 8     # @param head, a ListNode
 9     # @return a ListNode
10     def deleteDuplicates(self, head):
11         if head is None or head.next is None:
12             return head
13         
14         dummyhead = ListNode(0)
15         dummyhead.next = head
16         
17         p = dummyhead
18         while p.next is not None and p.next.next is not None:
19             tmp = p
20             while tmp.next.val == tmp.next.next.val:
21                 tmp = tmp.next
22                 if tmp.next.next is None:
23                     break
24             if tmp == p:
25                 p = p.next
26             else:
27                 if tmp.next.next is not None:
28                     p.next = tmp.next.next
29                 else:
30                     p.next = tmp.next.next
31                     break
32         return dummyhead.next