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