删除链接列表中的节点时出错<listnode.listnode object =“”at =“”0x0000000267320 =“”> </listnode.listnode>

时间:2015-03-04 17:41:37

标签: python linked-list

我正在创建一个链表,我必须为我的列表添加一些不同的函数,如max,min,count和index。我现在必须添加一个删除函数,这是一段代码。

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

我基本上只是尝试链接序列中的下一个节点,以便内置的垃圾清除器将从序列中删除该节点。但是,我收到以下错误消息,我知道必须对我的实例执行某些操作。

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>

处理完成,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

    #--------------------------------------------------------------

如果有人能帮助我找出为什么我的removeItem函数不起作用会有所帮助!

另外,我还试图制作这个列表的双向链表,我知道我需要在我的ListNode函数中添加一个prev_node函数,但还需要添加什么?再次感谢!

1 个答案:

答案 0 :(得分:2)

如果您的方法返回<LinkNode object at 0xmemoryaddr>字符串,那么它正在正常。您正在打印已删除的节点,并且Python正在使用该实例的默认repr()表示。

如果您想让它更具可读性,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)

这将打印ListNode(99, None)而不是<ListNode object at 0xmemoryaddr>字符串Python默认为:

>>> 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)

您必须考虑以下事项:您还需要调整列表的长度;成功删除后,从self.size中减去1。