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 nodestruct Node { int data; struct Node* next;}; // Function to find the second last// node of the linked listint 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 headvoid 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 codeint 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 nodeclass 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 traversalimport math # Link list nodeclass Node: def __init__(self, data): self.data = data self.next = None # Function to find the second last# node of the linked listdef 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 headdef push(head, new_data): new_node = Node(new_data) #new_node.data = new_data new_node.next = head head = new_node return head # Driver codeif __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 traversalusing System; // Linked list nodepublic 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 |
Output:
29
Time complexity : O(n)
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