Delete duplicate-value nodes from a sorted linked list


Problem Statement :


This challenge is part of a tutorial track by MyCodeSchool

You are given the pointer to the head node of a sorted linked list, where the data in the nodes is in ascending order. Delete nodes and return a sorted list with each distinct value in the original list. The given head pointer may be null indicating that the list is empty.

Example

head refers to the first node in the list 1 -> 2 -> 2- >  3 -> 3 -> 3 -> 3 -> NULL.

Remove 1 of the 2 data values and return  pointing to the revised list 1 -> 2 -> 3 -> NULL.

Function Description

Complete the removeDuplicates function in the editor below.

removeDuplicates has the following parameter:

SinglyLinkedListNode pointer head: a reference to the head of the list


Returns

SinglyLinkedListNode pointer: a reference to the head of the revised list


Input Format

The first line contains an integer t, the number of test cases.

The format for each test case is as follows:

The first line contains an integer n, the number of elements in the linked list.
Each of the next n lines contains an integer, the data value for each of the elements of the linked list.



Solution :



title-img


                            Solution in C :

In C++ :


/*
  Remove all duplicate elements from a sorted linked list
  Node is defined as 
  struct Node
  {
     int data;
     struct Node *next;
  }
*/
Node* RemoveDuplicates(Node *head)
{
  // This is a "method-only" submission. 
  // You only need to complete this method. 
    
    Node *ptr = head,*temp=NULL,*tmp=NULL;
    
    while(ptr!=NULL && ptr->next!=NULL)
    {
        temp = ptr->next;
        ptr->next=NULL;
        
        while(temp!=NULL && ptr->data == temp->data)
        {
            tmp=temp;
            temp=temp->next;
            
            tmp->next=NULL;
            delete(tmp);
        }
        ptr->next = temp;
        ptr = temp;
    }
    
    return head;
}



In Java : 


Node RemoveDuplicates(Node head) {
  // This is a "method-only" submission. 
  // You only need to complete this method. 
    
    if(head==null)
        return null;
    
    Node temp=head.next;
    Node prev=head;
    while(temp!=null)
        {
        
         if(prev.data==temp.data)
             {
             prev.next=temp.next;
             temp.next=null;
             temp=prev.next;
         }
        else
            {
              prev=temp;
              temp=temp.next;
        
        }
    }
    return head;
}



In python3 :


"""
 Delete duplicate nodes
 head could be None as well for empty list
 Node is defined as
 
 class Node(object):
 
   def __init__(self, data=None, next_node=None):
       self.data = data
       self.next = next_node

 return back the head of the linked list in the below method.
"""

def RemoveDuplicates(head):
    c = head
    p = None
    if c==None or c.next==None:
        return(head)
    else:
        p=c
        c=c.next
    while c!= None:
        if p.data==c.data:
            p.next = c.next
            c=c.next
        else:
            p = p.next
            c = c.next
    return(head)
  
  
  
  
  
  
  
 In C :


// Complete the removeDuplicates function below.

/*
 * For your reference:
 *
 * SinglyLinkedListNode {
 *     int data;
 *     SinglyLinkedListNode* next;
 * };
 *
 */
SinglyLinkedListNode* removeDuplicates(SinglyLinkedListNode* head) {
    struct SinglyLinkedListNode*temp1=head;
    struct SinglyLinkedListNode*temp=head->next;
    while(temp!=NULL)
    {
        if(head->data==temp->data)
        {
            temp=temp->next;
            head->next=temp;
        }
        else
        {
         head=temp;
         temp=temp->next;
        }
    }
 return temp1;

}
                        








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