题目要求：
Given a sorted linked list, delete all duplicates such that each element appear only once.
Examples：
Given 1->1->2, return 1->2.
Given 1->1->2->3->3, return 1->2->3.
解题思路：
- 指针移动示意图
注意else语句的链表相连操作！！！
代码：
class ListNode(object):
def __init__(self, x):
self.val = x
self.next = None
def __repr__(self):
if self:
return "{} -> {}".format(self.val, repr(self.next))
def my_print(self):
while self:
print(self.val)
self = self.next
class Solution(object):
def deleteDuplication(self, head):
"""
:param head: ListNode
:return: ListNode
"""
dummy = ListNode(-1)
dummy.next = head
prev = dummy
curr = head
while curr:
if prev.val == curr.val:
val = curr.val
while curr and curr.val == val:
curr = curr.next
prev.next = curr
else:
prev.next = curr
prev = curr
curr = curr.next
return dummy.next
if __name__ == "__main__":
head, head.next, head.next.next = ListNode(1), ListNode(2), ListNode(3)
head.next.next.next, head.next.next.next.next = ListNode(3), ListNode(4)
head.next.next.next.next.next, head.next.next.next.next.next.next = ListNode(4), ListNode(5)
print(Solution().deleteDuplication(head))
补充：
while curr and curr.val == val: 和 while curr.val == val: 相比较的话，最后curr变为None的时候没有val这个概念，所以应该先判断curr是否为None。
- 没有None.val这个玩意
