Write a GetNth() function that takes a linked list and an integer index and returns the data value stored in the node at that index position.
Example:
Input: 1->10->30->14, index = 2 Output: 30 The node at index 2 is 30
Algorithm:
1. Initialize count = 0 2. Loop through the link list a. If count is equal to the passed index then return current node b. Increment count c. change current to point to next of the current.
Implementation:
Python3
# A complete working Python program to # find n'th node in a linked list # Node class class Node:
# Function to initialize the node object
def __init__( self , data):
# Assign data
self .data = data
# Initialize next as null
self . next = None # Linked List class contains a Node object class LinkedList:
# Function to initialize head
def __init__( self ):
self .head = None
# This function is in LinkedList class.
# It inserts a new node at the beginning
# of Linked List.
def push( self , new_data):
# 1 & 2: Allocate the Node &
# Put in the data
new_node = Node(new_data)
# 3. Make next of new Node as head
new_node. next = self .head
# 4. Move the head to point to new Node
self .head = new_node
# Returns data at given index in linked list
def getNth( self , index):
# Initialise temp
current = self .head
# Index of current node
count = 0 # Loop while end of linked list
# is not reached
while (current):
if (count = = index):
return current.data
count + = 1
current = current. next
# If we get to this line, the caller was
# asking for a non-existent element so
# we assert fail
assert (false)
return 0
# Driver Code if __name__ = = '__main__' :
llist = LinkedList()
# Use push() to construct list
# 1->12->1->4->1
llist.push( 1 )
llist.push( 4 )
llist.push( 1 )
llist.push( 12 )
llist.push( 1 )
n = 3
print ( "Element at index 3 is :" ,
llist.getNth(n))
|
Output:
Element at index 3 is 4
Time Complexity: O(n)
Space Complexity: O(1) because using constant variables
Method 2- With Recursion:
Algorithm:
getnth(node,n) 1. Initialize count = 0 2. if count==n return node->data 3. else return getnth(node->next,n-1)
Implementation:
Python3
# Python3 program to find n'th node in # linked list using recursion class Node:
def __init__( self , data):
self .data = data
self . next = None
class LinkedList:
def __init__( self ):
self .head = None
# Given a reference (pointer to pointer) to
# the head of a list and an int, push a new
# node on the front of the list.
# Make new node and add
# into LinkedList
def push( self , new_data):
new_node = Node(new_data)
new_node. next = self .head
self .head = new_node
def getNth( self , llist, position):
# Call recursive method
llist.getNthNode( self .head,
position, llist)
# Recursive method to find Nth Node
def getNthNode( self , head, position, llist):
# Initialize count
count = 0 if (head):
# If count is equal to position,
# it means we have found the position
if count = = position:
print (head.data)
else :
llist.getNthNode(head. next ,
position - 1 , llist)
else :
# If head doesn't exist we have
# traversed the LinkedList
print ( 'Index Doesn\'t exist' )
# Driver Code if __name__ = = "__main__" :
llist = LinkedList()
llist.push( 1 )
llist.push( 4 )
llist.push( 1 )
llist.push( 12 )
llist.push( 1 )
# llist.getNth(llist,int(input()))
# Enter the node position here
# First argument is instance of LinkedList
print ( "Element at Index 3 is" , end = " " )
llist.getNth(llist, 3 )
# This code is contributed by Yogesh Joshi |
Output:
Element at index 3 is 4
Time Complexity: O(n)
Auxiliary Space: O(n) due to recursive call stack
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