C++ Program to create a Program for n’th node from the end of a Linked List or nth node end linked list.
To nicely understand this example to the nth node end linked list, you should have the knowledge of following C++ programming topics:

Method 1 (Use length of the linked list)
- Calculate the length of Linked List. Let the length be len.
- Print the (len – n + 1)th node from the beginning of the Linked List.
Program to find n’th node from the end
#include<stdio.h>
#include<stdlib.h>
/* Link list node */
struct Node
{
int data;
struct Node* next;
};
/* Function to get the nth node from the last of a linked list*/
void printNthFromLast(struct Node* head, int n)
{
int len = 0, i;
struct Node *temp = head;
// 1) count the number of nodes in Linked List
while (temp != NULL)
{
temp = temp->next;
len++;
}
// check if value of n is not more than length of the linked list
if (len < n)
return;
temp = head;
// 2) get the (n-len+1)th node from the begining
for (i = 1; i < len-n+1; i++)
temp = temp->next;
printf ("%d", temp->data);
return;
}
void push(struct Node** head_ref, int new_data)
{
/* allocate node */
struct Node* new_node =
(struct Node*) malloc(sizeof(struct Node));
/* put in the data */
new_node->data = new_data;
/* link the old list off the new node */
new_node->next = (*head_ref);
/* move the head to point to the new node */
(*head_ref) = new_node;
}
/* Drier program to test above function*/
int main()
{
/* Start with the empty list */
struct Node* head = NULL;
// create linked 35->15->4->20
push(&head, 20);
push(&head, 4);
push(&head, 15);
push(&head, 35);
printNthFromLast(head, 5);
return 0;
}
Output
35
recursive C code for the same method
void printNthFromLast(struct Node* head, int n)
{
static int i = 0;
if (head == NULL)
return;
printNthFromLast(head->next, n);
if (++i == n)
printf("%d", head->data);
}
Time Complexity: O(n) where n is the length of the linked list.
NEXT UP IN Link List
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