1

所以我已经解决了几个问题，通过在这里和我认识的人那里获得帮助。我的问题的根源是我不知道如何包装 None 以便我不会不断收到这些没有属性或不可调用的错误。

对于这个链表，我真正需要的是插入和打印列表。我没有包括打印列表，因为它很简单，并且不会引起问题。

错误位于 Linked_List 下、插入下、elif 下。它是这样评论的：#<----ERROR

这是代码：

class Node:
def __init__(self, word):
    self.data = word
    self.next = None
def nextNode(self):
    if self.next is not None:
        return self.next
    else:
        return None
def getData(self):
    return self.data
def setNext(self, node):
    self.next = node
def hasNext(self):
    if self.next == None:
        return False
    else:
        return True


class Linked_List:
def __init__(self):
    self.head = Node(None)
    self.isempty = True
def insert(self, word):
    newNode = Node(word)
    prev = self.head.nextNode()
    current = self.head.nextNode()
    nextFound = False #the next would be the current when it is less than node
    #Look for position to insert:

    #When empty
    if self.isempty == True:
        self.isempty = False
        self.head = newNode
    #When has more than one
    elif self.head.hasNext():
        while nextFound == False:
            if current.getData() > newNode.getData():
                prev = current
                current = curent.nextNode()
            else:
                nextFound = True
        #Insert
        prev.next().setNext(newNode) # <-------ERROR -----HERE~~
        newNode.setNext(current)
    else:
        #When only has one node not empty
        if self.head.getData() > newNode.getData():
            self.head.setNext(newNode)
        else:
            newNode.setNext(self.head)
            self.head = newNode

插入：

lList.insert(string)

在这里解决：

class Linked_List:
def __init__(self):
    self.head = Node(None)
    self.isempty = True
def insert(self, word):
    newNode = Node(word)
    prev = self.head.nextNode()
    current = self.head.nextNode()
    nextFound = False #the next would be the current when it is less than node
    #Look for position to insert:

    #When empty
    if self.isempty == True:
        self.isempty = False
        self.head = newNode
    #When has more than one
    elif self.head.hasNext():
        while nextFound == False and current != None:
            if current.getData() > newNode.getData():
                prev = current
                if current.hasNext():
                    current = current.nextNode()
                else:
                    current = None
            else:
                nextFound = True
        #Insert
        prev.setNext(newNode)
        newNode.setNext(current)
    else:
        #When only has one node not empty
        if self.head.getData() > newNode.getData():
            self.head.setNext(newNode)
        else:
            newNode.setNext(self.head)
            self.head = newNode
4

1 回答 1

0

你如何使用它？我猜你会做类似的事情yourList.insert(1)。在您的代码中，您执行以下操作：self.head = node，node用户传递给的内容在哪里insert。因此，在下一次打电话给insert您时，您最终会尝试调​​用 anint或您尝试放入列表中的任何内容。您需要将用户提供的任何对象与您的Node类一起包装：

def insert(self, thing):
    node = Node(thing)
    //...

但是，请记住发布所有相关代码，以便帮助您的人不必猜测。

编辑：仍然，编辑后的情况保持不变。你不包装传递给你的列表的对象，所以你一直试图调用Node非节点对象的方法......

于 2013-08-28T16:14:56.317 回答