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
Please refer complete article on Write a function to get Nth node in a Linked List for more details!
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