Question

所以我已经解决了几个问题，在这里获得帮助，并且来自我认识的人。我的问题的根源是我不知道如何包装无我没有得到这些没有属性或无法调用的错误。

对于这个链表，我真正需要的只是insert和printlist。我没有包含打印列表，因为它很简单，并且没有引起问题。

错误位于elif下插入下的Linked_List下。它评论如下：＃＆lt; ---- ERROR

以下是代码：

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)

解决方法：

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

Answer 1

你如何使用它？我猜你做的事情是yourList.insert(1)。在您的代码中，您执行：self.head = node，其中node是用户传递给insert的内容。因此，在下次拨打insert时，您最终会尝试拨打int或您尝试放入列表的任何内容。您需要使用Node类包装用户提供的任何对象：

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

但是，请记得发布所有相关代码，以便那些试图帮助您的人不必猜测。

编辑：仍然，编辑后案例仍然相同。你没有包装传递给你的列表的对象，所以你一直试图在非Node对象上调用Node方法......

（Python）'NoneType'对象不可调用（Linked List实现）

1 个答案: