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
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