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 listclass 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:
- Sort the elements using Merge Sort for Linked Lists.
- Remove duplicates in linear time using the algorithm for removing duplicates in sorted Linked List.
Below is the implementation for above approach:
Java
import java.io.*;// structure of a node in the linked listclass Node { int data; Node next; Node(int d) { data = d; next = null; }}// class for the linked listclass 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 classclass 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 linkedlistimport 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.
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