# Java Program For Removing All Occurrences Of Duplicates From A Sorted Linked List

• Last Updated : 10 Jan, 2022

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 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.

## Java

 `// Java program to remove all occurrences of``// duplicates from a sorted linked list `` ` `// class to create Linked lIst ``class` `LinkedList{``     ` `// Head of linked list ``Node head = ``null``; ``class` `Node``{``    ``// Value in the node ``    ``int` `val; ``    ``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;``    ``Node 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 code``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(``    ``"After 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)
Please refer complete article on Remove all occurrences of duplicates from a sorted Linked List for more details!

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