Python链接列表追加

时间:2015-02-04 06:53:00

标签: python linked-list append

我为一个UnorderedList()类创建了一个append方法,该类在我的IDLE窗口中正常工作,但当它被分配到大学的测试时:

my_list = UnorderedList()
my_list.append(13)
for num in my_list: 
    print(num, end=" ")
print()    

它会返回错误:AttributeError: Nonetype object has no attribute 'getNext'

这是追加方法:

def append(self,item):
    current = self.head
    while current.getNext() != None:
        current = current.getNext()
    current.setNext(Node(item))

以下是我的其他课程和代码:

class Node:
    def __init__(self,initdata):
        self.data = initdata
        self.next = None

    def getData(self):
        return self.data

    def getNext(self):
        return self.next

    def setData(self,newdata):
        self.data = newdata

    def setNext(self,newnext):
        self.next = newnext

class UnorderedList:

    def __init__(self):
        self.head = None
        self.count = 0

    def append(self,item):
        current = self.head
        while current.getNext() != None:
            current = current.getNext()
        current.setNext(Node(item))

为什么测试会返回该错误以及如何修复我的追加方法?

4 个答案:

答案 0 :(得分:7)

问题在于附加方法:

def append(self,item):
    current = self.head
    while current.getNext() != None:
        current = current.getNext()
    current.setNext(Node(item))

在第一次迭代中,current的值为self.head,最初设置为None,您不会检查它。

相反,改变这一点并在下面引入对此条件的检查:

def append(self,item):
    current = self.head
    if current:
        while current.getNext() != None:
            current = current.getNext()
        current.setNext(Node(item))
    else:
        self.head = Node(item)

PS:您还使用了一个变量self.count,您没有更新。您可能也希望同样更新它。

答案 1 :(得分:1)

你的append方法工作正常,但它遍历列表,直到找到最后一个节点 - 这使得它成为O(n)。如果跟踪最后一个节点,可以进行附加,即O(1):

> (if true 0)      ; test-expr = true, then-expr = 0
if: missing an "else" expression in: (if true 0)
> (if true 0 1)    ; test-expr = true, then-expr = 0, else-expr = 1
0
> (if false 0 1)   ; test-expr = false, then-expr = 0, else-expr = 1
1

要启用此功能,列表类构造函数应为:

def append_O1(self, item):
        temp = Node(item)
        last = self.tail
        last.setnext(temp)
        self.tail = temp
        self.length += 1

答案 2 :(得分:1)

添加更多精心设计的追加方法

在开头插入的方法

def inserAtBeginning(self, item):
    newNode = Node(item)
    newNode.setdata(item)
    if self.listLength() == 0:
        self.head = newNode
    else:
        newNode.setnext(self.head)
        self.head = newNode

最后插入的方法

def insertAtEnd(self, item):
    newNode = Node(item)
    newNode.setdata(item)
    current = self.head
    while current.getnext() != None:
        current = current.getnext()
    current.setnext(newNode)

在指定位置插入的方法

def insertAtPos(self, pos, item):
    if pos > self.listLength() or pos < 0:
        return None
    if pos == 0:
        self.inserAtBeginning(item)
    else:
        if pos == self.listLength():
            self.insertAtEnd(item)
        else:
            newNode = Node(item)
            newNode.setdata(item)
            current = self.head
            count = 0
            while count < pos - 1:
                count += 1
                current = current.getnext()
            newNode.setnext(current.getnext())
            current.setnext(newNode)

答案 3 :(得分:1)

此O(n)实现也应适用于空列表:

def append(self,item):
    current = self.head
    if current == None:
        self.head = Node(item)
    else:
        while current.getNext() != None:
            current = current.getNext()

        current.setNext(Node(item))
相关问题