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 = Noneclass 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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