python双向链表原理与实现方法详解

class Node(object):
  """双向链表节点"""
  def __init__(self, item):
    self.item = item
    self.next = None
    self.prev = None
class DLinkList(object):
  """双向链表"""
  def __init__(self):
    self.__head = None
  def is_empty(self):
    """判断链表是否为空"""
    return self.__head == None
  def length(self):
    """返回链表的长度"""
    cur = self.__head
    count = 0
    while cur != None:
      count += 1
      cur = cur.next
    return count
  def travel(self):
    """遍历链表"""
    cur = self.__head
    while cur != None:
      print cur.item,
      cur = cur.next
    print ""
  def add(self, item):
    """头部插入元素"""
    node = Node(item)
    if self.is_empty():
      # 如果是空链表，将_head指向node
      self.__head = node
    else:
      # 将node的next指向_head的头节点
      node.next = self.__head
      # 将_head的头节点的prev指向node
      self.__head.prev = node
      # 将_head 指向node
      self.__head = node
  def append(self, item):
    """尾部插入元素"""
    node = Node(item)
    if self.is_empty():
      # 如果是空链表，将_head指向node
      self.__head = node
    else:
      # 移动到链表尾部
      cur = self.__head
      while cur.next != None:
        cur = cur.next
      # 将尾节点cur的next指向node
      cur.next = node
      # 将node的prev指向cur
      node.prev = cur
  def search(self, item):
    """查找元素是否存在"""
    cur = self.__head
    while cur != None:
      if cur.item == item:
        return True
      cur = cur.next
    return False

  def insert(self, pos, item):
    """在指定位置添加节点"""
    if pos <= 0:
      self.add(item)
    elif pos > (self.length()-1):
      self.append(item)
    else:
      node = Node(item)
      cur = self.__head
      count = 0
      # 移动到指定位置的前一个位置
      while count < (pos-1):
        count += 1
        cur = cur.next
      # 将node的prev指向cur
      node.prev = cur
      # 将node的next指向cur的下一个节点
      node.next = cur.next
      # 将cur的下一个节点的prev指向node
      cur.next.prev = node
      # 将cur的next指向node
      cur.next = node

  def remove(self, item):
    """删除元素"""
    cur = self.__head
    while cur != None:
      # 找到了要删除的元素
      if cur.item == item:
        # 先判断此结点是否是头节点
        # 头节点
        if cur == self.__head:
          self.__head = cur.next
          # 如果存在下一个结点，则设置下一个结点
          if cur.next:
            # 判断链表是否只有一个结点
            cur.next.prev = None
        else:
          cur.prev.next = cur.next
          # 如果存在下一个结点，则设置下一个结点
          if cur.next:
            cur.next.prev = cur.prev
        break
      else:
        cur = cur.next