Given a sorted linked list, delete all nodes that have duplicate numbers (all occurrences),

leaving only numbers that appear once in the original list.

Examples:

Input : 23->28->28->35->49->49->53->53 Output : 23->35 Input : 11->11->11->11->75->75 Output : empty List

Note that this is different from Remove Duplicates From Linked List

The idea is to maintain a pointer* (prev)* to the node which just previous to the block of nodes we are checking for duplicates. In the first example, the pointer* prev* would point to 23 while we check for duplicates for the node 28. Once we reach the last duplicate node with value 28 (name it *current* pointer), we can make the next field of prev node to be the next of current and update *current=current.next*. This would delete the block of nodes with value 28 which has duplicates.

Below is Java implementation of the idea.

/* Java program to remove all occurrences of duplicates from a sorted linked list */ /* class to create Linked lIst */ class LinkedList { Node head=null; /* head of linked list */ class Node { int val; /* value in the node */ Node next; Node(int v) { /* default value of the next pointer field */ val = v; next = null; } } /* Function to insert data nodes into the Linked List at the front */ public void insert(int data) { Node new_node = new Node(data); new_node.next = head; head = new_node; } /* Function to remove all occurrences of duplicate elements */ public void removeAllDuplicates() { /* create a dummy node that acts like a fake head of list pointing to the original head*/ Node dummy = new Node(0); /* dummy node points to the original head*/ dummy.next = head; Node prev = dummy; current = head; while (current != null) { /* Until the current and previous values are same, keep updating current */ while (current.next != null && prev.next.val == current.next.val) current = current.next; /* if current has unique value i.e current is not updated, Move the prev pointer to next node*/ if (prev.next == current) prev = prev.next; /* when current is updated to the last duplicate value of that segment, make prev the next of current*/ else prev.next = current.next; current = current.next; } /* update original head to the next of dummy node */ head = dummy.next; } /* Function to print the list elements */ public void printList() { Node trav=head; if (head==null) System.out.print(" List is empty" ); while (trav != null) { System.out.print(trav.val + " "); trav = trav.next; } } /* Driver program to test above functions */ public static void main(String[] args) { LinkedList ll = new LinkedList(); ll.insert(53); ll.insert(53); ll.insert(49); ll.insert(49); ll.insert(35); ll.insert(28); ll.insert(28); ll.insert(23); System.out.println("Before removal of duplicates"); ll.printList(); ll.removeAllDuplicates(); System.out.println("\nAfter removal of duplicates"); ll.printList(); } }

Output:

List before removal of duplicates 23 28 28 35 49 49 53 53 List after removal of duplicates 23 35

Time Complexity : O(n)

This article is contributed by **Saloni Baweja**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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