Problem

Write a function that takes in the head node of a singly linked list and a value, and returns the node with the given value. If the value is not present in the linked list, return null.
Example input:
1 -> 2 -> 3 -> 4 -> null
3
Example output:
3

Solution

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

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

    def add_node(self, value):
        new_node = Node(value)
        if self.head is None:
            self.head = new_node
            return
        current_node = self.head
        while current_node.next is not None:
            current_node = current_node.next
        current_node.next = new_node

    def print_list(self):
        current_node = self.head
        while current_node is not None:
            print(current_node.value)
            current_node = current_node.next

    def find_node(self, value):
        current_node = self.head
        while current_node is not None:
            if current_node.value == value:
                return current_node
            current_node = current_node.next
        return None

# Test linked_list = LinkedList()
linked_list.add_node(1)
linked_list.add_node(2)
linked_list.add_node(3)
linked_list.add_node(4)

print(linked_list.find_node(3))
print(linked_list.find_node(5))



In Python write the optimal solution for the problem below. Add a comment explaining the solution and why it is optimal.

Problem:
Write a function that takes in the head node of a singly linked list and a value, and returns the node with the given value. If the value is not present in the linked list, return null.
Example input:
1 -> 2 -> 3 -> 4 -> null
3
Example output:
3