双向链接列表迭代器python

时间:2013-03-29 20:25:20

标签: python

我正在构建一个双向链表,我正在努力构建 PYTHON 中的双向链表迭代器方法。

这是我目前的代码

class DoubleListNode:
    def __init__(self,data):
        self.data=data
        self.prev = None
        self.next= None

class ListIterator:
    def __init__(self):
        self._current = self.head

    def __iter__(self):
        return self

    def next(self):
        if self.size == 0 :
            raise StopIteration
        else:
            item = self._current.data
            self._current=self._current.next
            return item

class DoublyLinkedList:
    def __init__(self):
        self.head= None
        self.tail= None
        self.size = 0

    def add(self,data):
        newnode= DoubleListNode(data)
        self.size+=1
        if self.head is None:
            self.head = newnode
            self.tail = self.head
        elif data < self.head.data: # before head
            newnode.next = self.head
            self.head.prev= newnode
            self.head= newnode
        elif data > self.tail.data: # at the end
            newnode.prev= self.tail
            self.tail.next= newnode
            self.tail=newnode
        else:
            curNode = self.head
            while curNode is not None and curNode.data < data:
                curNode=curNode.next            
            newnode.next= curNode
            newnode.prev=curNode.prev
            curNode.prev.next= newnode
            curNode.prev=newnode

    def remove(self,data):
        curNode=self.head
        while curNode is not None and curNode.data!= data:
            curNode= curNode.next
        if curNode is not None:
            self.size -= 1
            if curNode is self.head:
                self.head= curNode.next
            else:
                curNode.prev.next=curNode.next
            if curNode is self.tail:
                self.tail=curNode.prev
            else:
                curNode.next.prev=curNode.prev

当我进行测试时,它说TypeError: iteration over non-sequence。我做错了吗?

4 个答案:

答案 0 :(得分:3)

发布时,代码未初始化(即 self.head 未定义)。

但总的来说,你走在正确的轨道上。查看the source for Python's collections.OrderedDict以获取遍历双向链表的详尽示例。

这是一个简化的例子:

class Link:
    def __init__(self, value, prev=None, next=None):
        self.value = value
        self.prev = prev
        self.next = next

    def __iter__(self):
        here = self
        while here:
            yield here.value
            here = here.next

    def __reversed__(self):
        here = self
        while here:
            yield here.value
            here = here.prev

if __name__ == '__main__':
    a = Link('raymond')
    b = Link('rachel', prev=a);  a.next=b
    c = Link('matthew', prev=b); b.next=c

    print 'Forwards:'
    for name in a:
        print name
    print
    print 'Backwards:'
    for name in reversed(c):
        print name

答案 1 :(得分:2)

我认为有两件重要的事情需要解决。

首先,您的DoublyLinkedList课程没有__iter__方法。您可能想要创建一个返回ListIterator实例的实例。也许您正在尝试手动执行此操作,但这将是正常的方法。

其次,您需要修复ListIterator中的代码才能正常工作。目前,您的__init__方法未正确初始化,next方法会尝试访问不存在的size等成员变量。

这是我认为可行的实现:

def ListIterator(object):
    def __init__(self, node):
        self.current = node

    def __iter__(self):
        return self

    def next(self):
        if self.current is None:
            raise StopIteration()

        result = self.current.data
        self.current = self.current.next

        return result

class DoublyLinkedList(object):

    # all your current stuff, plus:

    def __iter__(self):
        return ListIterator(self.head)

作为旁注,在您当前的代码中,您定义的是没有基础的类。这在Python 3中很好(默认情况下object将是基础),但在Python 2中,这将导致获得“旧式”类。不推荐使用旧式类,并且您会发现某些语言功能无法正常使用它们(尽管根据我所知,迭代中没有涉及任何功能)。另一方面,如果您已经在使用Python 3,那么您需要在迭代器类中定义__next__方法,而不是next(没有下划线)。

答案 2 :(得分:1)

以下是Doubly Linked List类的示例。

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

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

    def insert(self, val):
        newNode = Node(val)
        if self.count == 0:
            self.head = newNode
            self.tail = newNode
        else:
            self.head.prev = newNode
            newNode.next = self.head
            self.head = newNode
        self.count += 1

    def insertToEnd(self, val):
        newNode = Node(val)
        if self.count == 0:
            self.head = newNode
            self.tail = newNode
        else:
            self.tail.next = newNode
            newNode.prev = self.tail
            self.tail = newNode
        self.count += 1

    def search(self, val):
        p = self.head
        while p is not None:
            if p.data == val:
                return p
            p = p.next

    def delete(self, val):
        curNode = self.head
        while curNode != None:
            if curNode.data == val:
                if curNode.prev != None:
                    curNode.prev.next = curNode.next
                else:
                    self.head = curNode.next

                if curNode.next != None:
                    curNode.next.prev = curNode.prev
                else:
                    self.tail = curNode.prev

                self.count -= 1

            curNode = curNode.next


    def show(self):
        s = ""
        p = self.head
        while p is not None:
            s += str(p.data) + ' ';
            p = p.next
        print(s + "| count: " + str(self.count))

答案 3 :(得分:0)

根据您发布的内容,我可以建议:

class ListIterator:
    # other stuff ...
    def __iter__(self):
        while self._current:
            yield self._current.data
            self._current = self._current.next
        self._current = self.head # Reset the current pointer

您不必实施next()

更新

以下是一个示例用法:

for data in myListIterator:
    print data

# Without reset, the second time around won't work:
for data in myListIterator:
    print data