我正在创建一个链表，我必须为列表添加一些不同的功能，例如max，min，count和index。现在，我必须添加一个remove函数，这是这段代码。
def removeItem(self, position):
''' removeItem removes a selected, because python has a built in "garbage remover",
you don't have to physically delete the node, you only have to skip that node link and python will destroy it
by it self.'''
currentNode = self.head
previousNode = None
count = 0
while count != position:
#This is a quick check to make sure the next node isn't empty.
if currentNode.link is None:
print("Position Invalid")
return None
previousNode = currentNode
currentNode = currentNode.link
count += 1
#Node.Link should link to the next node in the sequence.
previousNode.link = currentNode.link
return currentNode
我基本上只是想链接序列中的下一个节点，以便内置的垃圾清除程序将从序列中删除该节点。但是，我收到以下错误消息，我知道这与我的实例有关。
C:\Python33\python.exe "C:/Users/koopt_000/Desktop/College/Sophomore Semester 2/Computer Science 231/Chapter4/Test.py"
900
1
1
2
<ListNode.ListNode object at 0x0000000002679320>
流程结束，退出代码为0
为什么在最后打印出这个奇怪的ListNode.ListNode对象？
这是我的测试代码：
from ListNode import ListNode
from LinkedList import LinkedList
node1 = ListNode(1)
node2 = ListNode(900)
node3 = ListNode(3)
node4 = ListNode(99)
node1.link = node2
node2.link = node3
node3.link = node4
linked_list = LinkedList((1, 900, 3, 99))
print(linked_list.__max__())
print(linked_list.__min__())
print(linked_list.getCount(900))
print(linked_list.getIndex(3))
print(linked_list.removeItem(3))
这是我的ListNode类的代码：
# ListNode.py
class ListNode(object):
def __init__(self, item = None, link = None):
'''creates a ListNode with the specified data value and link
post: creates a ListNode with the specified data value and link'''
self.item = item
self.link = link
这是我的LinkedList类的代码：
from ListNode import ListNode
class LinkedList(object):
#--------------------------------------------------------------
def __init__(self, seq=()):
""" Pre: Creates a Linked List
Post: Creates a list containing the items in the seq=()"""
if seq == ():
# If there is no items to be put into the list, then it creates an empty one.
self.head = None
else:
# Creates a node for the first item.
self.head = ListNode(seq[0], None)
# If there are remaining items, then they're added while keeping track of the last node.
last = self.head
for item in seq[1:]:
last.link = ListNode(item, None)
last = last.link
self.size = len(seq)
#-------------------------------------------------------------
def __len__(self):
'''Pre: Nothing.
Post: Returns the number of items in the list.'''
return self.size
#-------------------------------------------------------------
def __max__(self):
''' Goes through each node and compares what the max is for the linked list.
Post: Finds the max of the linked list and returns that value.'''
if self.head is None:
return None
max_value = self.head.item
node = self.head.link
while node is not None:
if node.item > max_value:
max_value = node.item
node = node.link
return max_value
#--------------------------------------------------------------
def __min__(self):
''' Goes through each node and compares what the min is for the linked list.
Post: Finds the min of the linked list and returns that value.'''
if self.head is None:
return None
min_value = self.head.item
node = self.head.link
while node is not None:
if node.item < min_value:
min_value = node.item
node = node.link
return min_value
#--------------------------------------------------------------
def getCount(self, yourData):
''' This function counts the amount of times a certain item is in the Linked List.'''
count = 0
node = self.head
for i in range(self.size):
data = node.item
if data is yourData:
count += 1
node = node.link
return count
#--------------------------------------------------------------
def getIndex(self, yourData):
''' getIndex finds the index of the selected item and returns that value. '''
node = self.head
if node is None:
return None
for i in range(self.size):
data = node.item
if data == yourData:
return i
node = node.link
raise IndexError
#--------------------------------------------------------------
def removeItem(self, position):
''' removeItem removes a selected, because python has a built in "garbage remover",
you don't have to physically delete the node, you only have to skip that node link and python will destroy it
by it self.'''
currentNode = self.head
previousNode = None
count = 0
while count != position:
#This is a quick check to make sure the next node isn't empty.
if currentNode.link == None:
print("Position Invalid")
return None
previousNode = currentNode
currentNode = currentNode.link
count += 1
#Node.Link should link to the next node in the sequence.
previousNode.link = currentNode.link
return currentNode
#--------------------------------------------------------------
如果有人可以帮助我找出为什么我的removeItem函数不起作用，那将很有帮助！
顺便提一句，我也在尝试创建此列表的双向链接列表，我知道我需要在prevNode中添加一个prev_node函数，但是还需要添加什么呢？再次感谢！
最佳答案
如果您的方法返回一个<LinkNode object at 0xmemoryaddr>字符串，则它工作正常。您正在打印已删除的节点，Python正在对该实例使用默认的repr()表示形式。
如果要使其更具可读性，可以给ListNode一个object.__repr__ method：
def __repr__(self):
next = 'None' if not self.link else '...' # just to indicate
return 'ListNode({!r}, {})'.format(self.item, next)
然后，这将打印
ListNode(99, None)而不是Python默认为的<ListNode object at 0xmemoryaddr>字符串：>>> ll = LinkedList((1, 900, 3, 99))
>>> ll.head
ListNode(1, ...)
>>> ll.head.link
ListNode(900, ...)
>>> ll.head.link.link
ListNode(3, ...)
>>> ll.head.link.link.link
ListNode(99, None)
您必须要考虑的一件事：您也需要调整列表的长度。成功删除后，从
self.size减去1。关于python - 删除链接列表<ListNode.ListNode对象位于0x0000000267320>中的节点时出错，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/28861215/