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python实现数据结构中双向循环链表操作的示例

发布时间:2020-10-24 11:39:27 来源:脚本之家 阅读:149 作者:GUET_DM_WLB 栏目:开发技术

看此博客之前建议先看看B站的视频python数据结构与算法系列课程,该课程中未实现双向循环链表的操作,所以我按照该视频的链表思路实现了双向循环链表的操作,欢迎大家阅读与交流,如有侵权,请联系博主!

下面附上代码:

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


class DoubleCycleLinkList:
  def __init__(self, node=None):
    self.__head = node

  def is_empty(self):
    """判空"""
    if self.__head is None:
      return True
    return False

  def length(self):
    """链表长度"""
    if self.is_empty():
      return 0
    cur = self.__head
    count = 1
    while cur.next is not self.__head:
      count += 1
      cur = cur.next
    return count

  def travel(self):
    """遍历链表"""
    if self.is_empty():
      return
    cur = self.__head
    while cur.next is not self.__head:
      print(cur.elem, end=" ")
      cur = cur.next
    print(cur.elem, end=" ")
    print("")

  def add(self, elem):
    """头插法"""
    node = Node(elem)
    if self.is_empty():
      self.__head = node
      node.prev = node
      node.next = node
    else:
      self.__head.prev.next = node
      node.prev = self.__head.prev
      node.next = self.__head
      self.__head.prev = node
      self.__head = node

  def append(self, elem):
    """尾插法"""
    node = Node(elem)
    if self.is_empty():
      self.__head = node
      node.prev = node
      node.next = node
    else:
      node.next = self.__head
      node.prev = self.__head.prev
      self.__head.prev.next = node
      self.__head.prev = node

  def insert(self, pos, elem):
    """任一位置(pos)插入, 下标从0数起"""
    if pos <= 0:
      self.add(elem)
    elif pos > (self.length() - 1):
      self.append(elem)
    else:
      count = 0
      cur = self.__head
      node = Node(elem)
      while count < (pos - 1):
        count += 1
        cur = cur.next
      node.next = cur.next
      node.prev = cur
      node.next.prev = node
      cur.next = node

  def remove(self, elem):
    """删除某一节点,若有多个符合条件的节点,删除第一个即可"""
    if self.is_empty():
      return
    cur = self.__head
    while cur.next is not self.__head:
      if cur.elem == elem:
        if cur is self.__head:
          self.__head = cur.next
          cur.prev.next = cur.next
          cur.next.prev = cur.prev
        else:
          cur.prev.next = cur.next
          cur.next.prev = cur.prev
        break
      cur = cur.next
    if cur.elem == elem:
      cur.prev.next = self.__head
      self.head = cur.prev

  def search(self, elem):
    """查找某一个节点"""
    if self.is_empty():
      return False
    cur = self.__head
    while cur.next is not self.__head:
      if cur.elem == elem:
        return True
      cur = cur.next
    # while中处理不到尾节点,所以进行最后尾节点的判断
    if cur.elem == elem:
      return True
    return False

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