Find the second last node of a linked list in single traversal
Given a linked list. The task is to find the second last node of the linked list using a single traversal only.
Examples:
Input : List = 1 -> 2 -> 3 -> 4 -> 5 -> NULL
Output : 4Input : List = 2 -> 4 -> 6 -> 8 -> 33 -> 67 -> NULL
Output : 33
The idea is to traverse the linked list following the below approach:
- If the list is empty or contains less than 2 elements, return false.
- Otherwise, check if the current node is the second last node of the linked list or not. That is, if (current_node->next-next == NULL ) then the current node is the second last node.
- If the current node is the second last node, print the node otherwise move to the next node.
- Repeat the above two steps until the second last node is reached.
Below is the implementation of the above approach:
C++
// C++ program to find the second last node // of a linked list in single traversal #include <iostream> using namespace std; // Link list node struct Node { int data; struct Node* next; }; // Function to find the second last // node of the linked list int findSecondLastNode( struct Node* head) { struct Node* temp = head; // If the list is empty or contains less // than 2 nodes if (temp == NULL || temp->next == NULL) return -1; // Traverse the linked list while (temp != NULL) { // Check if the current node is the // second last node or not if (temp->next->next == NULL) return temp->data; // If not then move to the next node temp = temp->next; } } // Function to push node at head void push( struct Node** head_ref, int new_data) { Node* new_node = new Node; new_node->data = new_data; new_node->next = (*head_ref); (*head_ref) = new_node; } // Driver code int main() { /* Start with the empty list */ struct Node* head = NULL; /* Use push() function to construct the below list 8 -> 23 -> 11 -> 29 -> 12 */ push(&head, 12); push(&head, 29); push(&head, 11); push(&head, 23); push(&head, 8); cout << findSecondLastNode(head); return 0; } |
Java
// Java program to find the second last node // of a linked list in single traversal // Linked list node class Node { int data; Node next; Node( int d) { this .data = d; this .next = null ; } } class LinkedList { Node start; LinkedList() { start = null ; } // Function to push node at head public void push( int data) { if ( this .start == null ) { Node temp = new Node(data); this .start = temp; } else { Node temp = new Node(data); temp.next = this .start; this .start = temp; } } // method to find the second last // node of the linked list public int findSecondLastNode(Node ptr) { Node temp = ptr; // If the list is empty or contains less // than 2 nodes if (temp == null || temp.next == null ) return - 1 ; // This loop stops at second last node while (temp.next.next != null ) { temp = temp.next; } return temp.data; } // Driver code public static void main(String[] args) { LinkedList ll = new LinkedList(); /* Use push() function to construct the below list 8 -> 23 -> 11 -> 29 -> 12 */ ll.push( 12 ); ll.push( 29 ); ll.push( 11 ); ll.push( 23 ); ll.push( 8 ); System.out.println(ll.findSecondLastNode(ll.start)); } } // This code is Contributed by Adarsh_Verma |
Python3
# Python3 program to find the second last node # of a linked list in single traversal import math # Link list node class Node: def __init__( self , data): self .data = data self . next = None # Function to find the second last # node of the linked list def findSecondLastNode(head): temp = head # If the list is empty or # contains less than 2 nodes if (temp = = None or temp. next = = None ): return - 1 # Traverse the linked list while (temp ! = None ): # Check if the current node is the # second last node or not if (temp. next . next = = None ): return temp.data # If not then move to the next node temp = temp. next # Function to push node at head def push(head, new_data): new_node = Node(new_data) #new_node.data = new_data new_node. next = head head = new_node return head # Driver code if __name__ = = '__main__' : # Start with the empty list head = None # Use push() function to construct # the below list 8 . 23 . 11 . 29 . 12 head = push(head, 12 ) head = push(head, 29 ) head = push(head, 11 ) head = push(head, 23 ) head = push(head, 8 ) print (findSecondLastNode(head)) # This code is contributed by Srathore |
C#
// C# program to find the second last node // of a linked list in single traversal using System; // Linked list node public class Node { public int data; public Node next; public Node( int d) { this .data = d; this .next = null ; } } public class LinkedList { Node start; LinkedList() { start = null ; } // Function to push node at head public void push( int data) { if ( this .start == null ) { Node temp = new Node(data); this .start = temp; } else { Node temp = new Node(data); temp.next = this .start; this .start = temp; } } // method to find the second last // node of the linked list public int findSecondLastNode(Node ptr) { Node temp = ptr; // If the list is empty or contains less // than 2 nodes if (temp == null || temp.next == null ) return -1; // This loop stops at second last node while (temp.next.next != null ) { temp = temp.next; } return temp.data; } // Driver code public static void Main() { LinkedList ll = new LinkedList(); /* Use push() function to construct the below list 8 -> 23 -> 11 -> 29 -> 12 */ ll.push(12); ll.push(29); ll.push(11); ll.push(23); ll.push(8); Console.WriteLine(ll.findSecondLastNode(ll.start)); } } // This code is contributed by PrinciRaj1992 |
Javascript
<script> // Javascript program to find the second last node // of a linked list in single traversal // Link list node class Node { constructor(data) { this .data = data; this .next = null ; } } // method to find the second last // node of the linked list function findSecondLastNode( ptr) { var temp = head ; // If the list is empty or // contains less than 2 nodes if (temp == null || temp.next == null ) return -1 ; // Traverse the linked list while (temp != null ) { // Check if the current node is the // second last node or not if (temp.next.next == null ) return temp.data; // If not then move to the next node temp = temp.next; } } // Function to push node at head function push( data) { var new_node = new Node(data) ; new_node.next = head ; head = new_node ; return head ; } // Driver Code var head = null ; /* Use push() function to construct the below list 8 -> 23 -> 11 -> 29 -> 12 */ head = push(12); head = push(29); head = push(11); head = push(23); head = push(8); document.write(findSecondLastNode(head)); // This code is contributed by jana_sayantan. </script> |
Output
29
Time complexity: O(n), where N is the number of nodes in the LinkedList.
Auxiliary Space: O(1), as constant space is used.
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