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