Given a sorted linked list, delete all duplicates such that each element appear only once.

Example 1:

Input: 1->1->2
Output: 1->2

Example 2:

Input: 1->1->2->3->3
Output: 1->2->3 利用dummy来记住head,pre去记录上一个node,将head 和pre 比较(如果pre存在的话), 一旦相等,则删除掉head该点。注意,那个点实际上还是在,但是根据python的回收机制,没有指针指向该点,会被回收掉。
(也可以像code中那样,将head.next = None) Code
class ListNode:
def __init__(self, x):
self.val = x
self.next = None class Solution:
def removeDup(self, head):
dummy, pre = head, None
while head:
if pre and head.val == pre.val:
pre.next = head.next
head.next = None
else:
         pre = head
head = pre.next
return dummy

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