python实现数据结构中双向循环链表操作的示例
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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