python - DoubleLinkedList 获取内存位置而不是节点值 - Python

Question

我正在尝试创建一个双向链表，但我得到的是内存位置而不是节点值。感谢帮助

#!/usr/bin/python3

class Node:
  def __init__(self, num):
    self.next = None
    self.prev = None
    self.num = num

class DlinkedList:
   def __init__(self):
     self.start = None
     self.end = None
     self.size = 0 

   def addFirst(self, num):
     n = Node(num)

     if( self.start == None):
       self.start = n
       self.end = n
       print (self.start)
       print (self.end)
       print (self.size)
     else:
       n.next = self.start

d = DlinkedList()
print ("Add node 2")
d.addFirst(2)
print ("Add Node 1")
d.addFirst(1)

输出如下所示，这意味着我在这里做错了。

我知道代码还没有完成，但我正在尝试一点一点地开始

代码的输出是:

Add node 2
<__main__.Node object at 0x7f0c61219940>
<__main__.Node object at 0x7f0c61219940>
0
Add Node 1

Answer 1

看来您在这里遗漏了一些关键步骤。您需要确保增加计数器并将新节点链接到双向链表的前面。这是我最终得到的结果:

#!/usr/bin/python3

class Node:
  def __init__(self, num):
    self.next = None
    self.prev = None
    self.num = num

class DlinkedList:
   def __init__(self):
     self.start = None
     self.end = None
     self.size = 0 

   def addFirst(self, num):
     n = Node(num)

     if(self.start == None):
       self.start = n
       self.end = n

       print (self.start.num)
       print (self.end.num)
       print (self.size)
     else:
       n.next = self.start
       self.start.prev = n
       self.start = n
     self.size += 1

d = DlinkedList()
print ("Add node 2")
d.addFirst(2)
print ("Add Node 1")
d.addFirst(1)

print(d.start.num, d.start.next.num, d.end.prev.num, d.end.num)

iPython 的示例输出

Add node 2
2
2
0
Add Node 1
1 2 1 2
In [3]: d.addFirst(3)

In [4]: print(d.start.num, d.start.next.num, d.end.prev.num, d.end.num)
3 1 1 2