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

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.

## Python3

 `# Python3 implementation for the``# above approach` `# Creating node``class` `Node:``    ``def` `__init__(``self``, val):``        ``self``.val ``=` `val``        ``self``.``next` `=` `None``class` `LinkedList:``    ``def` `__init__(``self``):``        ``self``.head ``=` `None``        ` `    ``# Add node into beginning of linked list``    ``def` `push(``self``, new_data):``        ``new_node ``=` `Node(new_data)``        ``new_node.``next` `=` `self``.head``        ``self``.head ``=` `new_node``        ``return` `new_node``        ` `    ``# Function to remove all occurrences``    ``# of duplicate elements``    ``def` `removeAllDuplicates(``self``, temp):``        ` `        ``# temp is head node of linkedlist``        ``curr ``=` `temp` `        ``# print(' print something')``        ``head ``=` `prev ``=` `Node(``None``)``        ``head.``next` `=` `curr` `        ``# Here we use the same as we do in removing``        ``# duplicates and the only extra thing is that``        ``# we need to remove all elements``        ``# having duplicates that we did in 30-31``        ``while` `curr ``and` `curr.``next``:``            ` `            ``# until the current value and next value``            ``# are same keep updating the current value``            ``if` `curr.val ``=``=` `curr.``next``.val:``                ``while``(curr ``and` `curr.``next` `and``                      ``curr.val ``=``=` `curr.``next``.val):``                    ``curr ``=` `curr.``next``                    ` `                    ``# still one of duplicate values left``                    ``# so point prev to curr``                ``curr ``=` `curr.``next``                ``prev.``next` `=` `curr``            ``else``:``                ``prev ``=` `prev.``next``                ``curr ``=` `curr.``next``        ``return` `head.``next``        ` `    ``# for print the linkedlist``    ``def` `printList(``self``):``        ``temp1 ``=` `self``.head``        ``while` `temp1 ``is` `not` `None``:``            ``print``(temp1.val, end ``=` `" "``)``            ``temp1 ``=` `temp1.``next``            ` `    ``# For getting head of linkedlist``    ``def` `get_head(``self``):``        ``return` `self``.head` `# Driver Code``if` `__name__``=``=``'__main__'``:``    ``llist ``=` `LinkedList()``    ``llist.push(``53``)``    ``llist.push(``53``)``    ``llist.push(``49``)``    ``llist.push(``49``)``    ``llist.push(``35``)``    ``llist.push(``28``)``    ``llist.push(``28``)``    ``llist.push(``23``)``    ` `    ``print``(``'Created linked list is:'``)``    ``llist.printList()``    ``print``(``    ``'Linked list after deletion of duplicates:'``)``    ``head1 ``=` `llist.get_head()``    ``#print(head1)``    ``llist.removeAllDuplicates(head1)``    ``llist.printList()``        ` `# This code is contributed by PRAVEEN KUMAR(IIIT KALYANI)`

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!