python单链表反转出错到底怎么才能改对？

Question

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

#要支持的操作有1.is_empty()链表是否为空2.length()链表长度3.travel()遍历整个链表4.add(item)链表头部添加元素
#5.append(item)列表尾部添加元素6.insert(pos,item)指定位置添加元素7.remove(删除节点)8.search(item)查找节点是否存在

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

    def is_empty(self):
        return self.__head==None
    def length(self):
        cur = self.__head  
        count = 0
        while cur!= None: 
            cur = cur.next  
        return count
    def travel(self):
        cur=self.__head
        while cur!=None:
            print(cur.elem,end=' ')
            cur=cur.next
        print()

    def add(self,item):
        node=Node(item)
        node.next=self.__head
        self.__head=node
    def append(self,item):
        node=Node(item)
        if self.is_empty():
            self.__head=node
        else:
            cur=self.__head
            while cur.next!=None:
                cur=cur.next
            cur.next=node
    def insert(self,pos,item):
        if pos<=0:
            self.add(item)
        elif pos>self.length()-1:
            self.append(item)
        else:
            pre=self.__head
            count=0
            while count<pos-1:
                count+=1
                pre=pre.next

            node=Node(item)
            node.next=pre.next
            pre.next=node
    def remove(self,item):
        cur=self.__head
        pre=None
        while cur!=None:
            if cur.elem==item:
                if cur==self.__head:
                    self.__head=cur.next
                else:
                    pre.next=cur.next
                break
            else:
                pre=cur
                cur=cur.next
    def search(self,item):
        cur=self.__head
        while cur!=None:
            if cur.elem==item:
                return True
            else:
                cur=cur.next
        return False
    def reversLinkeList(self):#反转单链表
        cur=self.__head
        pre=None
        next=None
        while cur!=None:
            next=cur.next
            cur.next=pre
            pre=cur
            cur=next
        return pre

if __name__=="__main__":
    ll=SingleLinkList()
    print(ll.is_empty())
    print(ll.length())
    ll.append(1)
    ll.append(2)
    ll.add(8)
    ll.append(3)
    ll.append(4)
    ll.append(5)
    ll.append(6)
    ll.travel()#8123456
    ll.insert(2,4)
    ll.remove(5)
    ll.travel()#8142346
    print(ll.is_empty())
    print(ll.length())
    print()
    ll.reversLinkeList()
    ll.travel()#8

这个链表是 8142346 反转后应该是6432418 但是我的缺输出8 ，改了好久都没改对 求大大们帮帮忙，在原来的基础上看看应该怎么改才能输出正确的反转链表

Answer 1

class Node(object):
    
    def __init__(self,elem,next=None):
        self.elem=elem
        self.next=next
        
#要支持的操作有1.is_empty()链表是否为空2.length()链表长度3.travel()遍历整个链表4.add(item)链表头部添加元素
#5.append(item)列表尾部添加元素6.insert(pos,item)指定位置添加元素7.remove(删除节点)8.search(item)查找节点是否存在

class SingleLinkList(object):
    
    def __init__(self,node=None):
        self.__head=node
        
    def __str__(self):
        return f'List({self.__items__()})'
    
    def __items__(self):
        if self.is_empty():
            return None
        else:
            res,cur = [],self.__head
            while cur is not None:
                if cur.elem==None:break
                res.append(cur.elem)
                cur = cur.next
            return res
        
    def is_empty(self):
        return self.__head==None
    
    def length(self):
        cur = self.__head  
        count = 0
        while cur!= None:
            count += 1
            cur = cur.next
        return count
    def travel(self):
        cur=self.__head
        while cur!=None:
            if cur.elem==None:break;
            print(cur.elem,end=' ')
            cur=cur.next
        print()
        
    def add(self,item):
        node=Node(item)
        node.next=self.__head
        self.__head=node
        
    def append(self,item):
        node=Node(item)
        if self.is_empty():
            self.__head=node
        else:
            cur=self.__head
            while cur.next!=None:
                cur=cur.next
            cur.next=node
            
    def insert(self,pos,item):
        if pos<=0:
            self.add(item)
        elif pos>self.length()-1:
            self.append(item)
        else:
            pre=self.__head
            count=0
            while count<pos-1:
                count+=1
                pre=pre.next
            node=Node(item)
            node.next=pre.next
            pre.next=node
            
    def remove(self,item):
        cur=self.__head
        pre=None
        while cur!=None:
            if cur.elem==item:
                if cur==self.__head:
                    self.__head=cur.next
                else:
                    pre.next=cur.next
                break
            else:
                pre=cur
                cur=cur.next
                
    def search(self,item):
        cur=self.__head
        while cur!=None:
            if cur.elem==item:
                return True
            else:
                cur=cur.next
        return False
    
    def reversLinkeList(self):#反转单链表
        cur=self.__head
        pre=Node(None)
        while cur!=None:
            pre,cur = Node(cur.elem,pre),cur.next
        self.__head = pre
        return self
    
if __name__=="__main__":
    
    ll=SingleLinkList()
    print(ll.is_empty())
    print(ll.length())
    ll.append(1)
    ll.append(2)
    ll.add(8)
    ll.append(3)
    ll.append(4)
    ll.append(5)
    ll.append(6)
    ll.travel()#8123456
    ll.insert(2,4)
    ll.remove(5)
    ll.travel()#8142346
    print(ll.is_empty())
    print(ll.length())
    print(ll)
    ll.reversLinkeList()
    print(ll)
    ll.travel()#8

运行结果：

============================ RESTART: D:\test0.py ============================
True
0
8 1 2 3 4 5 6
8 1 4 2 3 4 6
False
7
List([8, 1, 4, 2, 3, 4, 6])
List([6, 4, 3, 2, 4, 1, 8])
6 4 3 2 4 1 8