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Q. Program to remove duplicate elements from a doubly linked list.

Explanation

In this program, we will create a doubly linked list and remove the duplicate, if present, by traversing through the list.

Original List:

Program to remove duplicate elements from a doubly linked list

List after removing duplicates:

Program to remove duplicate elements from a doubly linked list

In above list, node2 is repeated thrice, and node 3 is repeated twice. Current will point to head, and index will point to node next to current. Start traversing the list till a duplicate is found that is when current's data is equal to index's data. In above example, the first duplicate will be found at position 4. Assign index to another node temp. Connect index's previous node with index's next node. Delete temp which was pointing to duplicate node. This process will continue till all duplicates are removed.

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. removeDuplicateNode() will remove duplicate nodes from the list.
    1. Define a new node current which will initially point to head.
    2. Node index will always point to node next to current.
    3. Loop through the list until current points to null.
    4. Check whether current?s data is equal to index's data that means index is duplicate of current.
    5. Node temp will point to index to store duplicate node.
    6. The previous node to the index will point to next node to index.
    7. Since, temp is pointing to index, which is a duplicate node, so set temp to null.
  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:

Originals list: 
1 2 3 2 2 4 5 3 
List after removing duplicates: 
1 2 3 4 5 

C

Output:

Originals list: 
1 2 3 2 2 4 5 3 
List after removing duplicates: 
1 2 3 4 5 

JAVA

Output:

Originals list: 
1 2 3 2 2 4 5 3 
List after removing duplicates: 
1 2 3 4 5 

C#

Output:

Originals list: 
1 2 3 2 2 4 5 3 
List after removing duplicates: 
1 2 3 4 5 

PHP

Output:

Originals list: 
1 2 3 2 2 4 5 3 
List after removing duplicates: 
1 2 3 4 5 

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