Insert at end of Doubly Linked List -[Updated]

Write program to Insert at end of Doubly Linked List.

Problem Statement: Given a doubly-linked list, insert a node at the end of the doubly linked list.

Examples:

Example 1:
Input:  List : 1 <--> 2 <--> 3 <--> 4 <--> 5
	K :  6
Output: List :  1 <--> 2 <--> 3 <--> 4 <--> 5 <--> 6 
Explanation: the value given in k is inserted at the tail of the doubly linked lists.

Example 2:
Input:  List: NULL 
	K : 2
Output:	List : 2
Explanation: The list is empty so it will become the first node.

Solution

Disclaimer: Don’t jump directly to the solution, try it out yourself first.

Intuition:

To put at the end of the doubly linked list we can point the last node of a doubly linked list to the new node and the previous of the new node to the last node.

Approach:

To insert at the end of the linked list:

We will first create a new node.
We have to traverse to the end of the linked list and then change the next of the last node to a new node.
If the linked list is empty then make the new node as head of the linked list
Else traverse to the last node and change the last node then make the last node as previous of new node.

Code:

C++ Code

#include<bits/stdc++.h>

using namespace std;

class node {
  public:
    int data;
  node * next;
  node * prev;
  node(int val) {
    data = val;
    next = NULL;
    prev = NULL;
  }
};

void insertAtTail(node * & head, int val) {
  node * new_node = new node(val);
  if (head == NULL) {
    head = new_node;
    return;
  }
  node * temp = head;
  while (temp -> next != NULL) {
    temp = temp -> next;
  }
  temp -> next = new_node;
  new_node -> prev = temp;
}

void display(node * head) {
  node * temp = head;
  while (temp != NULL) {
    cout << temp -> data << "-->";
    temp = temp -> next;
  }
  cout << "NULL\n";
}

int main() {
  node* head=NULL;
  insertAtTail(head,1);
  insertAtTail(head,2);
  insertAtTail(head,3);
  insertAtTail(head,4);
  insertAtTail(head,5);
  display(head);

}

Output: 1–>2–>3–>4–>5–>NULL

Time Complexity: O(n) as, we are inserting in the end so we are traversing through the list

Space Complexity: O(1) As, we do not need extra space to store the node.

Java Code

import java.util.*;

class Node {
  
  int data;
  Node  next;
  Node  prev;
  Node(int val) {
    data = val;
    next = null;
    prev = null;
  }
}

class TUF{
static Node insertAtTail(Node head, int val) {
  Node  new_node = new Node(val);
  if (head == null) {
    head = new_node;
    return head;
  }
  Node  temp = head;
  while (temp . next != null) {
    temp = temp . next;
  }
  temp . next = new_node;
  new_node . prev = temp;
  return head;
}

static void display(Node  head) {
  Node  temp = head;
  while (temp != null) {
    System.out.print(temp.data+"-->");
    temp = temp . next;
  }
    System.out.println("NULL");
}

public static void main(String args[]) {
  Node head=null;
  head = insertAtTail(head,1);
  head = insertAtTail(head,2);
  head = insertAtTail(head,3);
  head = insertAtTail(head,4);
  head = insertAtTail(head,5);
  display(head);

}
}

Output: 1–>2–>3–>4–>5–>NULL

Time Complexity: O(n), as we are inserting in the end so we are traversing through the list

Space Complexity: O(1) , As we do not need extra space to store the node.

Special thanks to Ayush Pandey for contributing to this article on takeUforward. If you also wish to share your knowledge with the takeUforward fam, please check out this article