方法一思路：
转list去重，再新建链表返回结果。
注，本题结果要保留节点的原顺序
class Solution(object):
def deleteDuplicates(self, head):
"""
:type head: ListNode
:rtype: ListNode
"""
# 转list
tar = []
while head:
tar.append(head.val)
head = head.next
# 删除list中的重复元素且保留原序
ans = []
num = 0
for i in range(len(tar)):
if tar.count(tar[i]) == 1:
ans.append(tar[i])
num += 1
# 转链表返回
pre = node = ListNode(-1)
for j in range(num):
node.next = ListNode(0)
node.next.val = ans[j]
node = node.next
return pre.next
方法二思路：
用字典统计。
用例 [-3,-1,-1,0,0,0,0,0,2] 返回值是：[2,-3]，预期结果是：[-3,2]，没能保持原序，蓝瘦。
class Solution(object):
def deleteDuplicates3(self, head):
"""
:type head: ListNode
:rtype: ListNode
"""
if not head or not head.next:
return head
prev = head.next
while prev:
print("test:", prev.val)
prev = prev.next
dictval = {}
while head:
if head.val not in dictval:
dictval[head.val] = 1
else:
dictval[head.val] += 1
head = head.next
# 转链表返回
pre = node = ListNode(0)
for key, value in dictval.items():
if value == 1:
node.next = ListNode(0)
node.next.val = key
node = node.next
return pre.next
代码三：
class Solution(object):
def deleteDuplicates(self, head):
"""
:type head: ListNode
:rtype: ListNode
"""
ans = ListNode('a')
ans.next = head
pre, cur = None, ans
while cur:
pre = cur
cur = cur.next
while cur and cur.next and cur.val == cur.next.val:
temp = cur.val
while cur and cur.val == temp:
cur = cur.next
pre.next = cur
return ans.next