打印链表中的节点

Question

I'm very new to coding and I have a basic question about printing a node in a doubly linked list in Python.

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

    def __str__(self):
        return ('[%s][%d][%s]' % (repr(self.prev), self.key, repr(self.next)))

Obviously I then have some more code for the list class itself. Let's say I have a doubly linked list with two nodes:

node1 : key 21
node2 : key 10
head --> node1 <==> node2

If I do print(node1) I get:

[*location of prev node*][21][*location of next node*]

which works exactly how I want.

So 2 questions:

1. Is this code "messy" or acceptable syntax for the str method?

2. Instead of printing the location of the prev and next nodes, how would I print the names of the nodes instead eg [node7][82][node9] ?

Hope this makes sense, and thanks for any help!

Answer 1

Instead of having to create a self.name needed to be initialized for each Node . You could add a self.id initialized with a static counter incremented at each creation.

Step 1 - add a static counter and initialize the self.id attribute

The counter starts from 0 and is accessible through Node.counter .

class Node():
    counter = 0
    def __init__(self, key = None, prev = None, next = None):
        Node.counter += 1
        self.id = Node.counter
        self.key = key
        self.prev = prev
        self.next = next

Step 2 - modify the __str__ function to get the id of prev and next

The name of the self.prev and self.next is only displayed when not None .

Recover the name of the class with self.__class__.__name__ .

def __str__(self):
    sprev = repr(self.prev)
    if self.prev != None :
        sprev = '%s%d' % (self.__class__.__name__, self.prev.id)
    snext = repr(self.next)
    if self.next != None :
        snext = '%s%d' % (self.__class__.__name__, self.next.id)
    return ('[%s][%d][%s]' % (sprev, self.key, snext))

Here are some examples:

>>> node1 = Node(21)
>>> print(node1)
[None][21][None]
>>> node2 = Node(10,node1)
>>> print(node2)
[Node1][10][None]
>>> node3 = Node(11,node2,node1)
>>> print(node3)
[Node2][11][Node1]