Write a function that takes a list sorted in non-decreasing order and deletes any duplicate nodes from the list. The list should only be traversed once.
For example if the linked list is 11->11->11->21->43->43->60 then removeDuplicates() should convert the list to 11->21->43->60.
Algorithm:
Traverse the list from the head (or start) node. While traversing, compare each node with its next node. If the data of the next node is the same as the current node then delete the next node. Before we delete a node, we need to store the next pointer of the node
Implementation:
Functions other than removeDuplicates() are just to create a linked list and test removeDuplicates().
Java
// Java program to remove duplicates // from a sorted linked listclass LinkedList{ // Head of list Node head; // Linked list Node class Node { int data; Node next; Node(int d) { data = d; next = null; } } void removeDuplicates() { // Another reference to head Node curr = head; // Traverse list till the // last node while (curr != null) { Node temp = curr; /* Compare current node with the next node and keep on deleting them until it matches the current node data */ while(temp != null && temp.data == curr.data) { temp = temp.next; } /* Set current node next to the next different element denoted by temp */ curr.next = temp; curr = curr.next; } } // Utility functions // Inserts a new Node at front of // the list. public void push(int new_data) { /* 1 & 2: Allocate the Node & Put in the data*/ Node new_node = new Node(new_data); // 3. Make next of new Node as head new_node.next = head; // 4. Move the head to point to // new Node head = new_node; } // Function to print linked list void printList() { Node temp = head; while (temp != null) { System.out.print(temp.data+" "); temp = temp.next; } System.out.println(); } // Driver code public static void main(String args[]) { LinkedList llist = new LinkedList(); llist.push(20); llist.push(13); llist.push(13); llist.push(11); llist.push(11); llist.push(11); System.out.println( "List before removal of duplicates"); llist.printList(); llist.removeDuplicates(); System.out.println( "List after removal of elements"); llist.printList(); }} // This code is contributed by Rajat Mishra |
Output:
Linked list before duplicate removal 11 11 11 13 13 20 Linked list after duplicate removal 11 13 20
Time Complexity: O(n) where n is the number of nodes in the given linked list.
Auxiliary Space: O(1) , as there is no extra space used.
Recursive Approach :
Java
// Java Program to remove duplicates// from a sorted linked list class GFG{ // Link list node static class Node { int data; Node next; }; // The function removes duplicates // from a sorted list static Node removeDuplicates(Node head) { /* Pointer to store the pointer of a node to be deleted*/ Node to_free; // Do nothing if the list is empty if (head == null) return null; // Traverse the list till last node if (head.next != null) { // Compare head node with next node if (head.data == head.next.data) { /* The sequence of steps is important. to_free pointer stores the next of head pointer which is to be deleted.*/ to_free = head.next; head.next = head.next.next; removeDuplicates(head); } // This is tricky: only advance if no deletion else { removeDuplicates(head.next); } } return head; } // UTILITY FUNCTIONS /* Function to insert a node at the beginning of the linked list */ static Node push(Node head_ref, int new_data) { // Allocate node Node new_node = new Node(); // Put in the data new_node.data = new_data; // Link the old list of the // new node new_node.next = (head_ref); // Move the head to point to the // new node (head_ref) = new_node; return head_ref; } // Function to print nodes in a given // linked list static void printList(Node node) { while (node != null) { System.out.print(" " + node.data); node = node.next; } } // Driver code public static void main(String args[]) { // Start with the empty list Node head = null; /* Let us create a sorted linked list to test the functions. Created linked list will be 11.11.11.13.13.20 */ head = push(head, 20); head = push(head, 13); head = push(head, 13); head = push(head, 11); head = push(head, 11); head = push(head, 11); System.out.println("Linked list before" + " duplicate removal "); printList(head); // Remove duplicates from linked list head = removeDuplicates(head); System.out.println("Linked list after" + " duplicate removal "); printList(head); }} // This code is contributed by Arnab Kundu |
Output:
Linked list before duplicate removal 11 11 11 13 13 20 Linked list after duplicate removal 11 13 20
Time Complexity: O(n), where n is the number of nodes in the given linked list.
Auxiliary Space: O(n), due to recursive stack where n is the number of nodes in the given linked list.
Another Approach: Create a pointer that will point towards the first occurrence of every element and another pointer temp which will iterate to every element and when the value of the previous pointer is not equal to the temp pointer, we will set the pointer of the previous pointer to the first occurrence of another node.
Below is the implementation of the above approach:
Java
// Java program to remove duplicates// from a sorted linked listclass LinkedList{ // Head of list Node head; // Linked list Node class Node { int data; Node next; Node(int d) { data = d; next = null; } } // Function to remove duplicates // from the given linked list void removeDuplicates() { // Two references to head temp will // iterate to the whole Linked List // prev will point towards the first // occurrence of every element Node temp = head,prev = head; // Traverse list till the last node while (temp != null) { // Compare values of both pointers if(temp.data != prev.data) { /* if the value of prev is not equal to the value of temp that means there are no more occurrences of the prev data. So we can set the next of prev to the temp node.*/ prev.next = temp; prev = temp; } // Set the temp to the next node temp = temp.next; } /* This is the edge case if there are more than one occurrences of the last element */ if(prev != temp) { prev.next = null; } } // Utility functions // Inserts a new Node at front // of the list. public void push(int new_data) { /* 1 & 2: Allocate the Node & Put in the data */ Node new_node = new Node(new_data); // 3. Make next of new Node as head new_node.next = head; // 4. Move the head to point to // new Node head = new_node; } // Function to print linked list void printList() { Node temp = head; while (temp != null) { System.out.print(temp.data + " "); temp = temp.next; } System.out.println(); } // Driver code public static void main(String args[]) { LinkedList llist = new LinkedList(); llist.push(20); llist.push(13); llist.push(13); llist.push(11); llist.push(11); llist.push(11); System.out.print("List before "); System.out.println( "removal of duplicates"); llist.printList(); llist.removeDuplicates(); System.out.println( "List after removal of elements"); llist.printList(); }} // This code is contributed by Arshita |
Output:
List before removal of duplicates 11 11 11 13 13 20 List after removal of elements 11 13 20
Time Complexity: O(n) where n is the number of nodes in the given linked list.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Another Approach: Using Maps
The idea is to push all the values in a map and printing its keys.
Below is the implementation of the above approach:
Java
// Java program for the above approachimport java.io.*;import java.util.*;class Node{ int data; Node next; Node() { data = 0; next = null; }}class GFG { /* Function to insert a node at the beginning of the linked list */ static Node push(Node head_ref, int new_data) { // Allocate node Node new_node = new Node(); // Put in the data new_node.data = new_data; /* Link the old list of the new node */ new_node.next = (head_ref); /* Move the head to point to the new node */ head_ref = new_node; return head_ref; } /* Function to print nodes in a given linked list */ static void printList(Node node) { while (node != null) { System.out.print(node.data + " "); node = node.next; } } // Function to remove duplicates static void removeDuplicates(Node head) { HashMap<Integer, Boolean> track = new HashMap<>(); Node temp = head; while(temp != null) { if(!track.containsKey(temp.data)) { System.out.print(temp.data + " "); } track.put(temp.data, true); temp = temp.next; } } // Driver Code public static void main (String[] args) { Node head = null; /* Created linked list will be 11->11->11->13->13->20 */ head = push(head, 20); head = push(head, 13); head = push(head, 13); head = push(head, 11); head = push(head, 11); head = push(head, 11); System.out.print( "Linked list before duplicate removal "); printList(head); System.out.print( "Linked list after duplicate removal "); removeDuplicates(head); }}// This code is contributed by avanitrachhadiya2155 |
Output:
Linked list before duplicate removal 11 11 11 13 13 20 Linked list after duplicate removal 11 13 20
Time Complexity: O(Number of Nodes)
Space Complexity: O(Number of Nodes)
Please refer complete article on Remove duplicates from a sorted linked list for more details!
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