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Java Program For Removing Duplicates From An Unsorted Linked List

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Given an unsorted Linked List, the task is to remove duplicates from the list. 

Examples:

Input: linked_list = 12 -> 11 -> 12 -> 21 -> 41 -> 43 -> 21 
Output: 12 -> 11 -> 21 -> 41 -> 43 
Explanation: Second occurrence of 12 and 21 are removed.

Input: linked_list = 12 -> 11 -> 12 -> 21 -> 41 -> 43 -> 21 
Output: 12 -> 11 -> 21 -> 41 -> 43 

Naive Approach to Remove Duplicates from an Unsorted Linked List: 

The most simple approach to solve this, is to check each node for duplicate in the Linked List one by one. 

Below is the Implementation of the above approach:

Java




// Java program to remove duplicates from unsorted
// linked list
 
class LinkedList {
 
    static Node head;
 
    static class Node {
 
        int data;
        Node next;
 
        Node(int d)
        {
            data = d;
            next = null;
        }
    }
 
    /* Function to remove duplicates from an
       unsorted linked list */
    void remove_duplicates()
    {
        Node ptr1 = null, ptr2 = null, dup = null;
        ptr1 = head;
 
        /* Pick elements one by one */
        while (ptr1 != null && ptr1.next != null) {
            ptr2 = ptr1;
 
            /* Compare the picked element with rest
                of the elements */
            while (ptr2.next != null) {
 
                /* If duplicate then delete it */
                if (ptr1.data == ptr2.next.data) {
 
                    /* sequence of steps is important here
                     */
                    ptr2.next = ptr2.next.next;
                    System.gc();
                }
                else /* This is tricky */ {
                    ptr2 = ptr2.next;
                }
            }
            ptr1 = ptr1.next;
        }
    }
 
    void printList(Node node)
    {
        while (node != null) {
            System.out.print(node.data + " ");
            node = node.next;
        }
    }
 
    public static void main(String[] args)
    {
        LinkedList list = new LinkedList();
        list.head = new Node(10);
        list.head.next = new Node(12);
        list.head.next.next = new Node(11);
        list.head.next.next.next = new Node(11);
        list.head.next.next.next.next = new Node(12);
        list.head.next.next.next.next.next = new Node(11);
        list.head.next.next.next.next.next.next
            = new Node(10);
 
        System.out.println(
            "Linked List before removing duplicates : ");
        list.printList(head);
 
        list.remove_duplicates();
        System.out.println("\n");
        System.out.println(
            "Linked List after removing duplicates : ");
        list.printList(head);
    }
}
// This code has been contributed by Mayank Jaiswal


Output

Linked List before removing duplicates : 
10 12 11 11 12 11 10 

Linked List after removing duplicates : 
10 12 11 

Time Complexity: O(N2)
Auxiliary Space: O(1)

Remove duplicates from an Unsorted Linked List using Sorting:

Follow the below steps to Implement the idea:

Below is the implementation for above approach:

Java




import java.io.*;
 
// structure of a node in the linked list
class Node {
    int data;
    Node next;
 
    Node(int d)
    {
        data = d;
        next = null;
    }
}
 
// class for the linked list
class LinkedList {
    Node head;
 
    // function to insert a node in the linked list
    public void push(int new_data)
    {
        Node new_node = new Node(new_data);
        new_node.next = head;
        head = new_node;
    }
 
    // function to sort the linked list
    public void sortList()
    {
        // pointer to traverse the linked list
        Node current = head;
        Node index = null;
 
        // loop to traverse the linked list
        while (current != null) {
            // loop to compare current node with all other
            // nodes
            index = current.next;
            while (index != null) {
                // checking for duplicate values
                if (current.data == index.data) {
                    // deleting the duplicate node
                    current.next = index.next;
                    index = current.next;
                }
                else {
                    index = index.next;
                }
            }
            current = current.next;
        }
    }
 
    // function to display the linked list
    public void printList()
    {
        Node node = head;
        while (node != null) {
            System.out.print(node.data + " ");
            node = node.next;
        }
        System.out.println();
    }
}
 
// main class
class Main {
    public static void main(String[] args)
    {
        LinkedList ll = new LinkedList();
        ll.push(20);
        ll.push(13);
        ll.push(13);
        ll.push(11);
        ll.push(11);
        ll.push(11);
 
        System.out.println(
            "Linked List before removing duplicates : ");
        ll.printList();
 
        ll.sortList();
 
        System.out.println(
            "Linked List after removing duplicates : ");
        ll.printList();
    }
}


Output

Linked List before removing duplicates : 
11 11 11 13 13 20 
Linked List after removing duplicates : 
11 13 20 

Time Complexity: O(N log N)
Auxiliary Space: O(1)

Remove duplicates from an Unsorted Linked List using Hashing:

The idea for this approach is based on the following observations:

  • Traverse the link list from head to end.
  • For every newly encountered element, check whether if it is in the hash table:
    • if yes, remove it
    • otherwise put it in the hash table.
  • At the end, the Hash table will contain only the unique elements.

Below is the implementation of the above approach:

Java




// Java program to remove duplicates
// from unsorted linkedlist
 
import java.util.HashSet;
 
public class removeDuplicates {
    static class node {
        int val;
        node next;
 
        public node(int val) { this.val = val; }
    }
 
    /* Function to remove duplicates from a
       unsorted linked list */
    static void removeDuplicate(node head)
    {
        // Hash to store seen values
        HashSet<Integer> hs = new HashSet<>();
 
        /* Pick elements one by one */
        node current = head;
        node prev = null;
        while (current != null) {
            int curval = current.val;
 
            // If current value is seen before
            if (hs.contains(curval)) {
                prev.next = current.next;
            }
            else {
                hs.add(curval);
                prev = current;
            }
            current = current.next;
        }
    }
 
    /* Function to print nodes in a given linked list */
    static void printList(node head)
    {
        while (head != null) {
            System.out.print(head.val + " ");
            head = head.next;
        }
    }
 
    public static void main(String[] args)
    {
        /* The constructed linked list is:
         10->12->11->11->12->11->10*/
        node start = new node(10);
        start.next = new node(12);
        start.next.next = new node(11);
        start.next.next.next = new node(11);
        start.next.next.next.next = new node(12);
        start.next.next.next.next.next = new node(11);
        start.next.next.next.next.next.next = new node(10);
 
        System.out.println(
            "Linked list before removing duplicates :");
        printList(start);
 
        removeDuplicate(start);
 
        System.out.println(
            "\nLinked list after removing duplicates :");
        printList(start);
    }
}
 
// This code is contributed by Rishabh Mahrsee


Output

Linked list before removing duplicates :
10 12 11 11 12 11 10 
Linked list after removing duplicates :
10 12 11 

Time Complexity: O(N), on average (assuming that hash table access time is O(1) on average).  
Auxiliary Space: O(N), As extra space is used to store the elements in the stack.



Last Updated : 28 Jun, 2023
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