leetcode 【 Remove Duplicates from Sorted List II 】 python 实现

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

 # Definition for singly-linked list.
 # class ListNode:
 #     def __init__(self, x):
 #         self.val = x
 #         self.next = None
 
 class Solution:
     # @param head, a ListNode
     # @return a ListNode
     def deleteDuplicates(self, head):
         if head is None or head.next is None:
             return head
 
         dummyhead = ListNode(0)
         dummyhead.next = head
 
         p = dummyhead
         while p.next is not None and p.next.next is not None:
             tmp = p
             while tmp.next.val == tmp.next.next.val:
                 tmp = tmp.next
                 if tmp.next.next is None:
                     break
             if tmp == p:
                 p = p.next
             else:
                 if tmp.next.next is not None:
                     p.next = tmp.next.next
                 else:
                     p.next = tmp.next.next
                     break
         return dummyhead.next

思路：

设立虚表头 hummyhead 这样处理Linked List方便一些

p.next始终指向待比较的元素

while循环中再嵌套一个while循环，把重复元素都跳过去。

如果遇到了重复元素：p不动，p.next变化；如果没有遇到重复元素，则p=p.next

Tips: 使用指针之前 最好加一个逻辑判断 指针不为空