C++ program to get Nth node in a Linked List

In this tutorial, we are going to discuss how to get the Nth node in a singly linked list in C++.

Let us have a look at the examples given below.

Example:

Input : 10->15->20->25, index=3
Output : 25
The node at index 3 is 25.

Input : 11->12->13->14->15, index=2
Output : 13
The node at index 2 is 13.

Algorithm:

1.  Initialize a count variable, count=0
2. Traverse through the link list
3. if  count is equal to the current index then return the current node,
4. else increment the count and change the current to point to the immediate next.

 

Implementation: Find Nth node in a Linked List in C++

//C++ code to get the'N'th node in Linked List
#include<bits/stdc++.h>
#include<assert.h>
using namespace std;

//Linked List Node
struct node
{
      int data;
      struct node* next;
};

//function to create a new node
node *newnode(int data)   
{
     struct node *temp=new node;
     temp->data=data;
     temp->next=NULL;
     return temp;
}

//function that takes head of linked list and index
//and returns the data at that index
int getNth(node *head,int index)
{
//initializing the current node as the head
     node* current=head;

//initializing the count variable
     int count=0;

     while(current!=NULL)
     {
        if(count==index)
//returning the data at the given index
           return current->data;

//incrementing the count variable
           count++;
//incrementing the current next
           current=current->next;
     } 

//if the user is asking for a non-existent element so we assert fail
     assert(0);
}

//Dry run program to testify the getNth() function
int main()
{
//creating a linked list
   struct node* head=newnode(10);
   head->next=newnode(15);
   head->next->next=newnode(20);
   head->next->next->next=newnode(25);


 //checking the getNth() function
   cout<<"Element at index 3 is " << getNth(head,3);
   return 0;
}

 

Output:

Element at index 3 is 25

I hope this tutorial helps you understand how to get the Nth element in a linked list in a better way.