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.
C#
// C# program to remove duplicates from// unsorted linked listusing System;class List_ { Node head; class Node { public int data; public Node next; public 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 dup = ptr2.next; ptr2.next = ptr2.next.next; } else { ptr2 = ptr2.next; } } ptr1 = ptr1.next; } } void printList(Node node) { while (node != null) { Console.Write(node.data + " "); node = node.next; } } // Driver Code public static void Main(String[] args) { List_ list = new List_(); 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); Console.WriteLine( "Linked List_ before removing duplicates : "); list.printList(list.head); list.remove_duplicates(); Console.WriteLine(""); Console.WriteLine( "Linked List_ after removing duplicates : "); list.printList(list.head); }}// This code is contributed by gauravrajput1 |
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^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)
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.
C#
// C# program to remove duplicates// from unsorted linkedlistusing System;using System.Collections.Generic;class removeDuplicates{ class node { public int val; public 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<int> hs = new HashSet<int>(); // 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) { Console.Write(head.val + " "); head = head.next; }}// Driver codepublic 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); Console.WriteLine("Linked list before removing " + "duplicates :"); printList(start); removeDuplicate(start); Console.WriteLine("Linked list after removing " + "duplicates :"); printList(start);}}// This code is contributed by amal kumar choubey |
Output:
Linked list before removing duplicates: 10 12 11 11 12 11 10 Linked list after removing duplicates: 10 12 11
Thanks to bearwang for suggesting this method.
Time Complexity: O(n) on average (assuming that hash table access time is O(1) on average).
Please refer complete article on Remove duplicates from an unsorted linked list for more details!
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