双向链表的简单实现

双向链表的特点是：第一个元素的 prev 是None

# 双向链表：
class Node:
    def __init__(self,  node=None, prev=None,  next=None):
        # 注意顺序，因为实例化，一般不会关键字传参， 如果node= None(1) 1 是给node形参的，如果形参列表不同，则会出现异常
        self.prev = prev
        self.node = node
        self.next = next

    def __repr__(self):
        return '{} <-- {} --> {}'.format(
            self.prev.node if self.prev else None,
            self.node,
            None if self.next is None else self.next.node)

class DoubleLinkList:
    def __init__(self, head= None, tail=None):
        self.head = head
        self.tail = tail
        self.lens = 0

    def append(self, node):
        node = Node(node)
        if self.head is None:
            self.head = node
        else:
            self.tail.next = node
            node.prev = self.tail
        self.tail = node
        self.lens += 1

    def iternode(self, reverse=True):
        if self.head is None:
            raise  Exception('This is a Emppty linklist')
        current = self.head if reverse else self.tail
        while current:
            yield  current
            current = current.next if reverse else current.prev

    def __iter__(self):
        if self.head is None:
            raise  Exception('This is a Emppty linklist')
        current = self.head
        while current:
            yield current
            current = current.next

    def __getitem__(self, index):
        reverse = True if index>=0 else False
        start = 0 if index>=0 else 1
        for i, v in enumerate(ll.iternode(reverse), start):
            if i == abs(index):
                return v

    def __setitem__(self, index, value):
        ll[index].node = value

    def __len__(self):
        return self.len

    def remove(self):
        pass
        # 先判断是否为空
        # 需要遍历，利用iter
        # 注意是中间，第一个，还是尾部删除

    def pop(self):
        pass
        # 值删除尾部，self.tail.next = None

    def insert(self, index):
        pass
        # 类似于remove
    


ll = DoubleLinkList()
ll.append(1)
ll.append(2)
ll.append(3)
ll.append(4)
ll.append(5)

for i in ll.iternode(False):
    print(i)
print('----'* 20)
print(ll[-1])

ll[-1] = 10000
print(ll[-1])

 部分打印结果：