题目要求：
Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.
Eexamples：
Given 1->2->3->3->4->4->5, return 1->2->5.
Given 1->1->1->2->3, return 2->3.
解题思路：
- 指针移动示意图
需要遍历链表所有的元素，同时保证当前元素的后一个元素不为空，为空肯定就不会有元素和当前元素相同。
只要本元素和前一个元素相同就指针后移。
else语句开始的链表相连操作需要注意！！！
代码：
class ListNode:
def __init__(self, x):
self.val = x
self.next = None
def __repr__(self):
if self is None:
return "Nil"
else:
return "{} -> {}".format(self.val, repr(self.next))
class Solution(object):
def deleteDuplicates(self, head):
"""
:type head: ListNode
:rtype: ListNode
"""
dummy = ListNode(0)
pre, cur = dummy, head
while cur:
if cur.next and cur.next.val == cur.val:
val = cur.val;
while cur and cur.val == val:
cur = cur.next
pre.next = cur
else:
pre.next = cur
pre = cur
cur = cur.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().deleteDuplicates(head))
