# 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:

Returns

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 :

```                            ```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;
}
*/
{
// This is a "method-only" submission.
// You only need to complete this method.

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;
}

}

In Java :

// This is a "method-only" submission.
// You only need to complete this method.

return null;

while(temp!=null)
{

if(prev.data==temp.data)
{
prev.next=temp.next;
temp.next=null;
temp=prev.next;
}
else
{
prev=temp;
temp=temp.next;

}
}
}

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.
"""

p = None
if c==None or c.next==None:
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

In C :

// Complete the removeDuplicates function below.

/*
*
*     int data;
* };
*
*/
while(temp!=NULL)
{
{
temp=temp->next;
}
else
{
temp=temp->next;
}
}
return temp1;

}```
```

## Delete a Node

Delete the node at a given position in a linked list and return a reference to the head node. The head is at position 0. The list may be empty after you delete the node. In that case, return a null value. Example: list=0->1->2->3 position=2 After removing the node at position 2, list'= 0->1->-3. Function Description: Complete the deleteNode function in the editor below. deleteNo

## Print in Reverse

Given a pointer to the head of a singly-linked list, print each data value from the reversed list. If the given list is empty, do not print anything. Example head* refers to the linked list with data values 1->2->3->Null Print the following: 3 2 1 Function Description: Complete the reversePrint function in the editor below. reversePrint has the following parameters: Sing

Given the pointer to the head node of a linked list, change the next pointers of the nodes so that their order is reversed. The head pointer given may be null meaning that the initial list is empty. Example: head references the list 1->2->3->Null. Manipulate the next pointers of each node in place and return head, now referencing the head of the list 3->2->1->Null. Function Descriptio