How to print ith node in Linked List in Python
# Define a Node class
class Node:
def __init__(self, value, next=None):
self.value = value
self.next = next
# Define a LinkedList class
class LinkedList:
def __init__(self, head=None):
self.head = head
def get_node(self, index):
# Set the current node to the head
current = self.head
# Iterate through the list until the desired node is found
for i in range(index):
if current:
current = current.next
else:
raise IndexError('List index out of range')
# Return the value of the node
return current.value
# Test the LinkedList class
node1 = Node(1)
node2 = Node(2)
node3 = Node(3)
node4 = Node(4)
node1.next = node2
node2.next = node3
node3.next = node4
linked_list = LinkedList(node1)
print(linked_list.get_node(0)) # Output: 1
print(linked_list.get_node(2)) # Output: 3
This approach works by iterating through the list using a loop and returning the value of the ` i`-th node when it is found. If the index is out of range, an IndexError is raised
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