思路就是第一个循环拿到第一个node，第二个循环去和除了第一个node外的所有node对比，碰到值相等的node就进行指针跳过从而实现删除
实现函数：
def remove_duplicates(link_list):
outside = link_list.get_head()
while outside:
inside = outside
while inside:
if inside.get_next() is not None and outside.get_data() == inside.get_next().get_data():
inside.remove_after()
else:
inside = inside.get_next()
outside = outside.get_next()
例子：
list4 = LinkedList()
element_list = [10, 13, 13, 5, 13, 13, 42, 1, 5, 6, 8, 42, 42, 42, 13, 42]
for i in range(len(element_list)-1, -1, -1):
list4.add(element_list[i])
print(list4)
remove_duplicates(list4)
print(list4)
# 得到结果：
# [10, 13, 13, 5, 13, 13, 42, 1, 5, 6, 8, 42, 42, 42, 13, 42]
# [10, 13, 5, 42, 1, 6, 8]
Node和Linked list class：
class Node(object):
def __init__(self, data, next_node=None):
self.__data = data
self.__next = next_node
def get_data(self):
return self.__data
def set_data(self, data):
self.__data = data
def remove_after(self):
self.__next = self.__next.get_next()
class LinkedList:
def __init__(self):
self.__head = None
def get_head(self):
return self.__head
def add(self, item): # add to the beginning of the list
new_node = Node(item, self.__head)
self.__head = new_node
def __str__(self):
result_list = []
if self.__head is not None:
current = self.__head
while current is not None:
result_list.append(str(current.get_data()))
current = current.get_next()
return '[' + ', '.join(result_list) + ']'
