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Q. Program to create a doubly linked list of n nodes and display it in reverse order.

Explanation

In this program, we create a doubly linked list, then reverse the list by reversing the direction of the list and print out the nodes.

Program to create a doubly linked list of n nodes and display it in reverse order

Traverse through the list by swapping the previous pointer with next pointer of each node. Then, swap the position of head and tail node that is, head of the original list will become tail of new list and tail of the original list will become head of the new list. So, the reversed list will be:

Program to create a doubly linked list of n nodes and display it in reverse order

Algorithm

  1. Define a Node class which represents a node in the list. It will have three properties: data, previous which will point to the previous node and next which will point to the next node.
  2. Define another class for creating a doubly linked list, and it has two nodes: head and tail. Initially, head and tail will point to null.
  3. addNode() will add node to the list:
    1. It first checks whether the head is null, then it will insert the node as the head.
    2. Both head and tail will point to a newly added node.
    3. Head's previous pointer will point to null and tail's next pointer will point to null.
    4. If the head is not null, the new node will be inserted at the end of the list such that new node's previous pointer will point to tail.
    5. The new node will become the new tail. Tail's next pointer will point to null.
  4. reverse() will reverse the given doubly linked list.
    1. Define a node current which will initially point to head.
    2. Traverse through the list by making current to point to current.next in each iteration till current points to null.
    3. In each iteration, swap previous and next pointer of each node to reverse the direction of the list.
    4. In the end, swap the position of head and tail.
  5. display() will show all the nodes present in the list.
    1. Define a new node 'current' that will point to the head.
    2. Print current.data till current points to null.
    3. Current will point to the next node in the list in each iteration.

Solution

Python

Output:

Original List: 
1 2 3 4 5 
Reversed List: 
5 4 3 2 1 

C

Output:

Original List: 
1 2 3 4 5 
Reversed List: 
5 4 3 2 1 

JAVA

Output:

Original List: 
1 2 3 4 5 
Reversed List: 
5 4 3 2 1 

C#

Output:

Original List: 
1 2 3 4 5 
Reversed List: 
5 4 3 2 1 

PHP

Output:

Original List: 
1 2 3 4 5 
Reversed List: 
5 4 3 2 1 

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