 GeeksforGeeks App
Open App Browser
Continue

# Python Program For Removing Duplicates From An Unsorted Linked List

Write a removeDuplicates() function that takes a list and deletes any duplicate nodes from the list. The list is not sorted.
For example if the linked list is 12->11->12->21->41->43->21 then removeDuplicates() should convert the list to 12->11->21->41->43.

METHOD 1 (Using two loops):
This is the simple way where two loops are used. Outer loop is used to pick the elements one by one and the inner loop compares the picked element with the rest of the elements.
Thanks to Gaurav Saxena for his help in writing this code.

## Python3

 `# Python3 program to remove duplicates``# from unsorted linked list``class` `Node():   ``    ``def` `__init__(``self``, data):       ``        ``self``.data ``=` `data``        ``self``.``next` `=` `None` `class` `LinkedList():   ``    ``def` `__init__(``self``):``        ` `        ``# Head of list``        ``self``.head ``=` `None` `    ``def` `remove_duplicates(``self``):       ``        ``ptr1 ``=` `None``        ``ptr2 ``=` `None``        ``dup ``=` `None``        ``ptr1 ``=` `self``.head` `        ``# Pick elements one by one``        ``while` `(ptr1 !``=` `None` `and``               ``ptr1.``next` `!``=` `None``):``            ` `            ``ptr2 ``=` `ptr1` `            ``# Compare the picked element with``            ``# rest of the elements``            ``while` `(ptr2.``next` `!``=` `None``):``                ` `                ``# If duplicate then delete it``                ``if` `(ptr1.data ``=``=` `ptr2.``next``.data):``                    ` `                    ``# Sequence of steps is important here``                    ``dup ``=` `ptr2.``next``                    ``ptr2.``next` `=` `ptr2.``next``.``next``                ``else``:``                    ``ptr2 ``=` `ptr2.``next``                    ` `            ``ptr1 ``=` `ptr1.``next``            ` `    ``# Function to print nodes in a``    ``# given linked list``    ``def` `printList(``self``):``        ``temp ``=` `self``.head``        ` `        ``while``(temp !``=` `None``):``            ``print``(temp.data, end ``=` `" "``)``            ``temp ``=` `temp.``next``            ` `        ``print``()``        ` `# Driver code``list` `=` `LinkedList()``list``.head ``=` `Node(``10``)``list``.head.``next` `=` `Node(``12``)``list``.head.``next``.``next` `=` `Node(``11``)``list``.head.``next``.``next``.``next` `=` `Node(``11``)``list``.head.``next``.``next``.``next``.``next` `=` `Node(``12``)``list``.head.``next``.``next``.``next``.``next``.``next` `=` `Node(``11``)``list``.head.``next``.``next``.``next``.``next``.``next``.``next` `=` `Node(``10``)` `print``(``"Linked List before removing duplicates :"``)``list``.printList()``list``.remove_duplicates()``print``()``print``(``"Linked List after removing duplicates :"``)``list``.printList()``# This code is contributed by maheshwaripiyush9`

Output:

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

Time Complexity: O(n^2)
Auxiliary Space: O(1)

METHOD 2 (Use Sorting):
In general, Merge Sort is the best-suited sorting algorithm for sorting linked lists efficiently.
1) Sort the elements using Merge Sort. We will soon be writing a post about sorting a linked list. O(nLogn)
2) Remove duplicates in linear time using the algorithm for removing duplicates in sorted Linked List. O(n)
Please note that this method doesn’t preserve the original order of elements.
Time Complexity: O(nLogn)

Auxiliary Space :O(1)

METHOD 3 (Use Hashing):
We traverse the link list from head to end. For every newly encountered element, we check whether it is in the hash table: if yes, we remove it; otherwise we put it in the hash table.

## Python3

 `# Python3 program to remove duplicates``# from unsorted linkedlist``class` `Node:   ``    ``def` `__init__(``self``, data):       ``        ``self``.data ``=` `data``        ``self``.``next` `=` `None` `class` `LinkedList:   ``    ``def` `__init__(``self``):       ``        ``self``.head ``=` `None``        ` `    ``# Function to print nodes in a ``    ``# given linked list``    ``def` `printlist(``self``):       ``        ``temp ``=` `self``.head``        ` `        ``while` `(temp):``            ``print``(temp.data, end ``=` `" "``)``            ``temp ``=` `temp.``next``            ` `    ``# Function to remove duplicates from a``    ``# unsorted linked list``    ``def` `removeDuplicates(``self``, head):``        ` `        ``# Base case of empty list or``        ``# list with only one element``        ``if` `self``.head ``is` `None` `or` `self``.head.``next` `is` `None``:``            ``return` `head``            ` `        ``# Hash to store seen values``        ``hash` `=` `set``() ` `        ``current ``=` `head``        ``hash``.add(``self``.head.data)` `        ``while` `current.``next` `is` `not` `None``:``            ``if` `current.``next``.data ``in` `hash``:``                ``current.``next` `=` `current.``next``.``next``            ``else``:``                ``hash``.add(current.``next``.data)``                ``current ``=` `current.``next` `        ``return` `head` `# Driver code``if` `__name__ ``=``=` `"__main__"``:``    ` `    ``# Creating Empty list``    ``llist ``=` `LinkedList()``    ``llist.head ``=` `Node(``10``)``    ``second ``=` `Node(``12``)``    ``third ``=` `Node(``11``)``    ``fourth ``=` `Node(``11``)``    ``fifth ``=` `Node(``12``)``    ``sixth ``=` `Node(``11``)``    ``seventh ``=` `Node(``10``)``    ` `    ``# Connecting second and third``    ``llist.head.``next` `=` `second``    ``second.``next` `=` `third``    ``third.``next` `=` `fourth``    ``fourth.``next` `=` `fifth``    ``fifth.``next` `=` `sixth``    ``sixth.``next` `=` `seventh` `    ``# Printing data``    ``print``(``"Linked List before removing Duplicates."``)``    ``llist.printlist()``    ``llist.removeDuplicates(llist.head)``    ``print``(``"Linked List after removing duplicates."``)``    ``llist.printlist()``# This code is contributed by rajataro0`

Output:

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