One-way circular linked list implementation in Python

Introduction

The so-called one-way circular list, but on the basis of a one-way list, such as Rattlesnake-like to connect its end and end, there are many similarities and must pay attention to the point. In particular, it may involve the operation of the head end node, not negligence.

What are the conditions for a lot of the traversal that must be done? And where should we stop?

What to do if the node to be deleted is the head or tail node when the delete operation is done? If the list has only one node, what should I do with it?

Code implementation

classNode (object):def __init__(self, value):#element FieldsSelf.value =value#link DomainSelf.next =NoneclassCircularlinkedlistoneway (object):def __init__(Self, node=None):#when constructing a non-empty chain, let its address field point to itself        ifNode:node.next=node self.__head=nodedefIs_empty (self):#The head node is not a none and is not empty        returnSelf.__head==Nonedef __len__(self): Count= 1cur= self.__head        ifself.is_empty ():return0 whileCur.next! = self.__head: Count+ = 1cur=Cur.nextreturnCountdefTraversal (self):ifself.is_empty ():returncur= self.__head         whileCur.next! = self.__head:            Print(cur.value) cur=Cur.next#cur is the tail node when exiting a loop        Print(Cur.value)defAdd (self, value):"""Head Insertion Method"""node=Node (value)ifSelf.is_empty (): Self.__head=node self.__head. Next = self.__head            returncur= self.__head         whileCur.next! = self.__head: Cur=Cur.next#the next pointer to the new node points to the original head nodeNode.next = self.__head        #point the new node to the head nodeSelf.__head=node#tail node next pointer to new head nodeCur.next = self.__head    defAppend (self, value): Node=Node (value) cur= self.__head        ifSelf.is_empty (): Self.__head=node self.__head. Next = self.__head            return         whileCur.next! = self.__head: Cur=Cur.next Node.next=Cur.next Cur.next=nodedefInsert (self, POS, value):ifPOS <=0:self.add (value)elifPOS > Len (self)-1: Self.append (value)Else: Node=Node (value) cur= self.__headCount=0 whileCount < Pos-1: Count+ = 1cur=Cur.next Node.next=Cur.next Cur.next=nodedefSearch (self, value):ifself.is_empty ():returnFalse cur= self.__head         whileCur.next! = self.__head:            ifCur.value = =Value:returnTrueElse: Cur=Cur.next#don't forget the tail node outside the while loop        ifCur.value = =Value:returnTruereturnFalsedefRemove (self, value): Cur= self.__headPrior=Noneifself.is_empty ():return         whileCur.next! = self.__head:            #node to delete if found            ifCur.value = =Value:#the node to be deleted is in the head                ifCur = = self.__head: Rear= self.__head                     whileRear.next! = self.__head: Rear=Rear.next self.__head=Cur.next Rear.next= self.__head                #the node to be deleted is in the middle                Else: Prior.next=Cur.next
# This is not a break to jump out of the loop, but a return of the Exit function, because it's done.return            #If you haven't found it yet ,            Else: Prior=cur cur=Cur.next#the node to be deleted at the tail        ifCur.value = =Value:#If there is only one element in the list, then the prior is None,next attribute will be error            #just make the head element none at this point            ifCur = = self.__head: Self.__head=NonereturnPrior.next= Cur.next

One-way circular linked list implementation in Python