python - 帮助使用 Python 中的循环链表

Question

我正在尝试制作一个循环单向链表。我希望能够针对单个喜欢的列表修改我的代码，但我遇到了一些麻烦。

对于我的链接列表，我有:

class Link (object):
  def __init__ (self, data, next = None):
    self.data = data
    self.next = next


class LinkedList(object):
  def __init__(self):
    self.first = None

  def __str__(self):
    a = "["
    current = self.first
    while current != None:
      a += str(current.data) + ', ' 
      current = current.next
    a = a[:-2] + ']'  
    return a  

  def __iter__(self):
    current = self.first
    a = []
    while current != None:
      a += [current.data]
      current = current.next
    return iter(a)

  def __len__ (self):
    current = self.first
    a = []
    while current != None:
      a += [current.data]
      current = current.next
    return len(a)

  def InsertFirst(self, item):
    NewLink = Link(item, self.first)
    self.first = NewLink

  def InsertLast(self, item):
    NewLink = Link(item)
    current = self.first

    if current == None:
      self.first = NewLink  
      return 

    while current.next != None:
      current = current.next
    current.next = NewLink 

  def Search(self, item):
    count = 0
    current = self.first
    while current != None:
      count += 1
      if current.data == item:
        return count
      else:
        pass
        current = current.next
    return -1

  def Delete(self, item):
    current = self.first
    previous = self.first

    if (current == None):
      return None

    while (current.data != item):
      if (current.next == None):
        return None
      else:
        previous = current
        current = current.next

    if (current == self.first):
      self.first = self.first.next
    else:
      previous.next = current.next

    return current

到目前为止，对于我的循环列表，我有:

class Link (object):
  def __init__ (self, data, next = None):
    self.data = data
    self.next = next


class CircularList(object):
  def __init__(self):
    self.first = Link(None, None)
    self.head = Link(None, self.first)

  def __str__(self):
    a = "["
    current = self.first
    while current != None:
      a += str(current.data) + ', ' 
      current = current.next
    a = a[:-2] + ']'  
    return a  

  def InsertLast(self, item):
    NewLink = Link(item)
    current = self.first

    if current == None:
      self.first = NewLink  
      return 

    while current.next != None:
      current = current.next
    current.next = Link(item)

我的问题是如何将最后一个元素链接回第一个元素以便横向移动？

Answer 1

循环链表的要点是跳过所有“if next is not None”逻辑。一开始，head指向自己，说明链表是空的。无需创建一个空的“第一个”——在一开始就做:

self.head = Link(None, None)
self.head.next = self.head

然后要在其他节点之后插入一个节点，您只需执行以下操作:

def insert_after(insert_node, after_node):
    insert_node.next = after_node.next
    after_node.next = insert_node

要在列表的开头插入，请执行:

insert_after(node, head)

插入之前需要迭代以找到“之前”节点，因为列表仅是单链接的:

def insert_before(node, before_node):
    loc = head
    while loc.next is not before_node:
        loc = loc.next
    insert_after(insert_node, loc)

要在列表末尾插入，请执行:

insert_before(node, head)

要获取列表中的所有元素，请执行以下操作:

current = self.head.next
while current is not self.head:
    # do something with current.data

    # advance to next element
    current = current.next

但循环列表的真正威力在于使其具有双向链接，因此您可以在不迭代的情况下插入。