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Python

Posted by syalih
syalih
16,468 Points

Reverse LinkedList?

Hi, can anyone please briefly explain me this below code what is exactly going on actually I don't get it from "for value in linked_list" because Node' object is not iterable" then how it is working in this function?

def reverse(linked_list):
        """
    Reverse the inputted linked list

    Args:
       linked_list(obj): Linked List to be reversed
    Returns:
       obj: Reveresed Linked List
    """
    new_list = LinkedList()
    node = linked_list.head
    prev_node = None

    for value in linked_list:
        new_node = Node(value)
        new_node.next = prev_node
        prev_node = new_node
    new_list.head = prev_node
    return new_list

llist = LinkedList()
for value in [4,2,5,1,-3,0]:
    llist.append(value)

flipped = reverse(llist)
is_correct = list(flipped) == list([0,-3,1,5,2,4]) and list(llist) == list(reverse(flipped))
print("Pass" if is_correct else "Fail")

2 Answers

Jeff Muday
MOD
Jeff Muday
Treehouse Moderator 28,720 Points
Jeff Muday
Jeff Muday
Treehouse Moderator 28,720 Points

The key to the iterable is the __iter__() and that it must use a generator (Python yield keyword).

Another way to make a linked list is to use Python list() as a parent class. Though, in some respects, this defeats the point of the exercise to make a LinkedList() that would follow operation of the classic data structure.

In my current career, which is in academics and research, I rarely, if ever, work with Linked Lists. Not because these structures aren't fundamental, but because these are abstracted in libraries like NumPy, SciPy, TensorFlow, etc. These are in the libraries but never directly used by the application programmer. This is especially true of modern languages (Python, R, Ruby, Java) which have automatic garbage collection. So in some ways the concept of LinkedList() as a fundamental design pattern in these languages is antiquated. In the same way learning about assembly and machine language is antiquated. We use lists, dictionaries, objects and don't have to worry about allocation or disposal.

Why Learn about Linked Lists?

The Linked List is a rich structure used to build up other structures like (abstract) Stacks, Queues, Double-Linked Lists, Trees, and so on. These structures are ever-present in the "stack" and "framework", but as modern programmers and developers we are focused on getting applications into use or to market and don't have to re-invent the wheel!

syalih
syalih
16,468 Points

Ohh now I totally get it thanks.

Jeff Muday
MOD
Jeff Muday
Treehouse Moderator 28,720 Points
Jeff Muday
Jeff Muday
Treehouse Moderator 28,720 Points

Without seeing the definition of the class LinkedList() it is impossible to tell. Here is a link that has a nice treatment of Linked Lists in Python.

https://stackabuse.com/python-linked-lists/

My guess is that your definition of LinkedList() class DOES NOT HAVE an __iter()__ method.

If that is the case, it will not yield values and cannot be used like a Python iterable (in a for loop). There are some linked lists which use a next() method or getter.

https://www.geeksforgeeks.org/getter-and-setter-in-python/

syalih
syalih
16,468 Points

here is whole code please check and thanks for the resource for LinkedList :) 

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

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

    def append(self, value):
        if self.head is None:
            self.head = Node(value)
            return

        node = self.head
        while node.next:
            node = node.next

        node.next = Node(value)

    def __iter__(self):
        node = self.head
        while node:
            yield node.value
            node = node.next

    def __repr__(self):
        return str([v for v in self])

def reverse(linked_list):
    new_list = LinkedList()
    node = linked_list.head
    prev_node = None

    # A bit of a complex operation here. We want to take the
    # node from the original linked list and prepend it to 
    # the new linked list
    for value in linked_list:
        new_node = Node(value)
        new_node.next = prev_node
        prev_node = new_node
    new_list.head = prev_node
    return new_list

llist = LinkedList()
for value in [4,2,5,1,-3,0]:
    llist.append(value)

flipped = reverse(llist)
is_correct = list(flipped) == list([0,-3,1,5,2,4]) and list(llist) == list(reverse(flipped))
print("Pass" if is_correct else "Fail")