将 LinkedList 类转换为 Circular Linked List 类的最简单方法是什么？

Question

我有一个包含近 200 行代码的 LinkedList 类。我想通过始终确保任何myLL.tail.next is myLL.head 来创建一个新的class LLCircular(LinkedList)。我相信我需要相应地更新append()、push()、remove() 等。有没有办法让原始的 LinkedList 类保持完整？也许是一个装饰器或一些笨拙的方法？

为简洁起见，如果阅读代码，我的push() 方法与append() 正好相反。我还有一个 pop() 和 remove() 方法，如果我只是重写这些方法，就需要更新它们。由于我试图避免这种方法，因此我没有发布那部分代码。

class LinkedListNode:
   def __init__(self, value, nextNode=None, prevNode=None):
      self.value  = value
      self.next   = nextNode
      self.prev   = prevNode
   def __str__(self):
      return str(self.value)

class LinkedList:
   def __init__(self, values=None):
      self.head   = None
      self.tail   = None
      if values is not None:
         self.append(values)
   def __str__(self):
      values = [str(x) for x in self]
      return ' -> '.join(values)
   def append(self, value=None):
      if value is None:
         raise ValueError('ERROR: LinkedList.py: append() `value` PARAMETER MISSING')
      if isinstance(value, list):
         for v in value:
            self.append(v)
         return
      elif self.head is None:
         self.head   = LinkedListNode(value)
         self.tail   = self.head
      else:
         ''' We have existing nodes  '''
         ''' Head.next is same '''
         ''' Tail is new node '''
         self.tail.next = LinkedListNode(value, None, self.tail)
         self.tail      = self.tail.next
         if self.head.next is None:
            self.head.next = self.tail.prev
      return self.tail

'''class LLCircular(LinkedList):'''
    ''' ??? '''

测试代码：

foo = LinkedList([1,2,3])
foo.tail.next = foo.head #My LL is now circular
cur = foo.head
i = 0
while cur:
   print(cur)
   cur = cur.next
   i +=1
   if i>9:
      break

Answer 1

您想要的是使用super 关键字调用您的LinkedList 基类函数，然后对LLCircular 类函数进行轻微修改，即：

class LLCircular(LinkedList):
    def append(self, value=None):
        super(LLCircular, self).append(value)
        # In addition to having called the LinkedList append, now you want
        # to make sure the tail is pointing at the head
        self.tail.next = self.head
        self.head.prev = self.tail

Answer 2

如果它是“圆形”的，它就不需要尾巴或头，对吗？ “追加”也没有意义——insert_after 和 insert_before 方法就足够了——此外，任何节点都是对完整循环列表的引用，不需要不同的对象：

class Circular:
    def __init__(self, value=None):
        self.value = value
        self.next = self
        self.previous = self
    def insert_after(self, value):
        node = Circular(value)
        node.next = self.next
        node.previous = self
        self.next.previous = node
        self.next = node
    def insert_before(self, value):
        node = Circular(value)
        node.next = self
        node.previous = self.previous
        self.previous.next = node
        self.previous = node
    def del_next(self):
        self.next = self.next.next
        self.next.previous = self
    def __iter__(self):
        cursor = self.next
        yield self
        while cursor != self:
             yield cursor
             cursor = cursor.next
    def __len__(self):
        return sum(1 for _ in self)