Pls explain this code to me

Question

class node:
def __init__(self, data = None, next = None):
self.data = data
self.next = next

class linked_list:
def __init__(self):
self.head = None

# function to add a node at front
def add_at_front(self, data):
self.head = node(data=data, next=self.head)

# function to check whether the list is empty
def is_empty(self):
return self.head == None

# function to add node at the end
def add_at_end(self, data):
if not self.head:
self.head = node(data=data)
return
curr = self.head
while curr.next:
curr = curr.next
curr.next = node(data=data)

# function to delete any node
def delete_node(self, key):
curr = self.head
prev = None
while curr and curr.data != key:
prev = curr
curr = curr.next
if prev is None:
self.head = curr.next
elif curr:
prev.next = curr.next

Answer 1

what are you having trouble understanding?

Answer 2

That method places a new node instance in the first spot (not the last one like usual.) Then it makes the first node(from before) become the 2nd node in line when connected to the new node. Does this make sense?

2-3-4-5
Then if I add 1 at front
1-2-3-4-5

Answer 3

The add_at_front method

Answer 4

Can you give me any other kind of example referring to this code:
self.head = node(data=data, next=self.head)

Answer 5

i have a linked list, lets call it lett_ls:

b-c-d-e

it has an attribute called .head

lett_ls.head.data == 'b'
the above statement is true.

the first thing add_to_front() does is make a new instance of a node.
It then loads up with data and assigns the previous head to be the new nodes next attribute(the linking of linked lists).

lets lett_ls.add_to_front('a')...
a-b-c-d-e

Answer 6

Well in your linked list "b-c-d-e" which one is node?

Answer 7

what you see is a representation of it. Its all the data of the nodes "linked" by - characters. Its for visual

Answer 8

Its all in your head, just depends on how you see it. Thats how i like to picture linked lists

Answer 9

Will you explain me the add_at_end and delete_node methods

Answer 10

yeah shoot me a message with the code and id be more than happy to