Write a function to get Nth node in a Linked List
Write a GetNth() function that takes a linked list and an integer index and returns the data value stored in the node at that index position.
Example:
Input: 1->10->30->14, index = 2
Output: 30
The node at index 2 is 30
Program for n’th node from the end of a Linked List
Given a Linked List and a number n, write a function that returns the value at the n’th node from the end of the Linked List.
For example, if the input is below list and n = 3, then output is “B”
Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.
Method 1 (Use length of linked list)
1) Calculate the length of Linked List. Let the length be len.
2) Print the (len – n + 1)th node from the beginning of the Linked List.
Double pointer concept : First pointer is used to store the address of the variable and second pointer used to store the address of the first pointer. If we wish to change the value of a variable by a function, we pass pointer to it. And if we wish to change value of a pointer (i. e., it should start pointing to something else), we pass pointer to a pointer.
Below is the implementation of the above approach:
C++14
|
// Simple C++ program to find n'th node from end
#include
using namespace std;
/* 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;
// 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;
// get the (len-n+1)th node from the beginning
for (i = 1; i < len - n + 1; i++)
temp = temp->next;
cout << temp->data;
return;
}
void push(struct Node** head_ref, int new_data)
{
/* allocate node */
struct Node* new_node = new 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;
}
// Driver Code
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, 4);
return 0;
}
|
Java
|
// Simple Java program to find n'th node from end of linked list
class LinkedList {
Node head; // head of the list
/* Linked List node */
class Node {
int data;
Node next;
Node(int d)
{
data = d;
next = null;
}
}
/* Function to get the nth node from the last of a
linked list */
void printNthFromLast(int n)
{
int len = 0;
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 (len-n+1)th node from the beginning
for (int i = 1; i < len - n + 1; i++)
temp = temp.next;
System.out.println(temp.data);
}
/* 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;
}
/*Driver program to test above methods */
public static void main(String[] args)
{
LinkedList llist = new LinkedList();
llist.push(20);
llist.push(4);
llist.push(15);
llist.push(35);
llist.printNthFromLast(4);
}
} // This code is contributed by Rajat Mishra
|
Python3
|
# Simple Python3 program to find
# n'th node from end
class Node:
def __init__(self, new_data):
self.data = new_data
self.next = None
class LinkedList:
def __init__(self):
self.head = None
# createNode and make linked list
def push(self, new_data):
new_node = Node(new_data)
new_node.next = self.head
self.head = new_node
# Function to get the nth node from
# the last of a linked list
def printNthFromLast(self, n):
temp = self.head # used temp variable
length = 0
while temp is not None:
temp = temp.next
length += 1
# print count
if n > length: # if entered location is greater
# than length of linked list
print('Location is greater than the' +
' length of LinkedList')
return
temp = self.head
for i in range(0, length - n):
temp = temp.next
print(temp.data)
# Driver Code
llist = LinkedList()
llist.push(20)
llist.push(4)
llist.push(15)
llist.push(35)
llist.printNthFromLast(4)
# This code is contributed by Yogesh Joshi
|
C#
|
// C# program to find n'th node from end of linked list
using System;
public class LinkedList
{
public Node head; // head of the list
/* Linked List node */
public class Node
{
public int data;
public Node next;
public Node(int d)
{
data = d;
next = null;
}
}
/* Function to get the nth node from the last of a
linked list */
void printNthFromLast(int n)
{
int len = 0;
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 (len-n+1)th node from the beginning
for (int i = 1; i < len - n + 1; i++)
temp = temp.next;
Console.WriteLine(temp.data);
}
/* 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;
}
/*Driver code */
public static void Main(String[] args)
{
LinkedList llist = new LinkedList();
llist.push(20);
llist.push(4);
llist.push(15);
llist.push(35);
llist.printNthFromLast(4);
}
}
// This code is contributed by Rajput-Ji
|
Javascript
Output
35
Following is a recursive C code for the same method. Thanks to Anuj Bansal for providing following code.
C++
|
void printNthFromLast(struct Node* head, int n)
{
int i = 0;
if (head == NULL)
return;
printNthFromLast(head->next, n);
if (++i == n)
cout<data;
}
|
C
|
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);
}
|
Java
|
static void printNthFromLast(Node head, int n)
{
int i = 0;
if (head == null)
return;
printNthFromLast(head.next, n);
if (++i == n)
System.out.print(head.data);
}
// This code is contributed by rutvik_56.
|
Python3
|
def printNthFromLast(head, n):
i = 0
if (head == None)
return
printNthFromLast(head.next, n);
i+=1
if (i == n):
print(head.data)
# This code is contributed by sunils0ni.
|
C#
|
static void printNthFromLast(Node head, int n)
{
static int i = 0;
if (head == null)
return;
printNthFromLast(head.next, n);
if (++i == n)
Console.Write(head.data);
}
// This code is contributed by pratham76.
|
Javascript
Time Complexity: O(n) where n is the length of linked list.
Method 2 (Use two pointers)
Maintain two pointers – reference pointer and main pointer. Initialize both reference and main pointers to head. First, move the reference pointer to n nodes from head. Now move both pointers one by one until the reference pointer reaches the end. Now the main pointer will point to nth node from the end. Return the main pointer.
Below image is a dry run of the above approach:
Below is the implementation of the above approach:
C++
|
// C++ program to find n-th node
// from the end of the linked list.
#include
using namespace std;
struct node {
int data;
node* next;
node(int val)
{
data = val;
next = NULL;
}
};
struct llist {
node* head;
llist() { head = NULL; }
// insert operation at the beginning of the list.
void insertAtBegin(int val)
{
node* newNode = new node(val);
newNode->next = head;
head = newNode;
}
// finding n-th node from the end.
void nthFromEnd(int n)
{
// create two pointers main_ptr and ref_ptr
// initially pointing to head.
node* main_ptr = head;
node* ref_ptr = head;
// if list is empty, return
if (head == NULL) {
cout << "List is empty" << endl;
return;
}
// move ref_ptr to the n-th node from beginning.
for (int i = 1; i < n; i++) {
ref_ptr = ref_ptr->next;
if (ref_ptr == NULL) {
cout << n
<< " is greater than no. of nodes in "
"the list"
<< endl;
return;
}
}
// move ref_ptr and main_ptr by one node until
// ref_ptr reaches end of the list.
while (ref_ptr != NULL && ref_ptr->next != NULL) {
ref_ptr = ref_ptr->next;
main_ptr = main_ptr->next;
}
cout << "Node no. " << n
<< " from end is: " << main_ptr->data << endl;
}
void displaylist()
{
node* temp = head;
while (temp != NULL) {
cout << temp->data << "->";
temp = temp->next;
}
cout << "NULL" << endl;
}
};
int main()
{
llist ll;
for (int i = 60; i >= 10; i -= 10)
ll.insertAtBegin(i);
ll.displaylist();
for (int i = 1; i <= 7; i++)
ll.nthFromEnd(i);
return 0;
}
// This code is contributed by sandeepkrsuman.
|
Java
|
// Java program to find n'th
// node from end using slow and
// fast pointers
class LinkedList
{
Node head; // head of the list
/* Linked List node */
class Node {
int data;
Node next;
Node(int d)
{
data = d;
next = null;
}
}
/* Function to get the
nth node from end of list */
void printNthFromLast(int n)
{
Node main_ptr = head;
Node ref_ptr = head;
int count = 0;
if (head != null)
{
while (count < n)
{
if (ref_ptr == null)
{
System.out.println(n
+ " is greater than the no "
+ " of nodes in the list");
return;
}
ref_ptr = ref_ptr.next;
count++;
}
if(ref_ptr == null)
{
if(head != null)
System.out.println("Node no. " + n +
" from last is " +
head.data);
}
else
{
while (ref_ptr != null)
{
main_ptr = main_ptr.next;
ref_ptr = ref_ptr.next;
}
System.out.println("Node no. " + n +
" from last is " +
main_ptr.data);
}
}
}
/* 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;
}
/*Driver program to test above methods */
public static void main(String[] args)
{
LinkedList llist = new LinkedList();
llist.push(20);
llist.push(4);
llist.push(15);
llist.push(35);
llist.printNthFromLast(4);
}
}
// This code is contributed by Rajat Mishra
|
Python3
|
# Python program to find n'th node from end using slow
# and fast pointer
# Node class
class Node:
# Constructor to initialize the node object
def __init__(self, data):
self.data = data
self.next = None
class LinkedList:
# Function to initialize head
def __init__(self):
self.head = None
# Function to insert a new node at the beginning
def push(self, new_data):
new_node = Node(new_data)
new_node.next = self.head
self.head = new_node
def printNthFromLast(self, n):
main_ptr = self.head
ref_ptr = self.head
count = 0
if(self.head is not None):
while(count < n ):
if(ref_ptr is None):
print ("% d is greater than the no. pf nodes in list" %(n))
return
ref_ptr = ref_ptr.next
count += 1
if(ref_ptr is None):
self.head = self.head.next
if(self.head is not None):
print("Node no. % d from last is % d "
%(n, main_ptr.data))
else:
while(ref_ptr is not None):
main_ptr = main_ptr.next
ref_ptr = ref_ptr.next
print ("Node no. % d from last is % d "
%(n, main_ptr.data))
if __name__ == '__main__':
llist = LinkedList()
llist.push(20)
llist.push(4)
llist.push(15)
llist.push(35)
llist.printNthFromLast(4)
# This code is contributed by Nikhil Kumar Singh(nickzuck_007)
|
C#
|
// C# program to find n'th node from end using slow and
// fast pointerspublic
using System;
public class LinkedList
{
Node head; // head of the list
/* Linked List node */
public class Node
{
public int data;
public Node next;
public Node(int d)
{
data = d;
next = null;
}
}
/* Function to get the nth node from end of list */
void printNthFromLast(int n)
{
Node main_ptr = head;
Node ref_ptr = head;
int count = 0;
if (head != null)
{
while (count < n)
{
if (ref_ptr == null)
{
Console.WriteLine(n + " is greater than the no "
+ " of nodes in the list");
return;
}
ref_ptr = ref_ptr.next;
count++;
}
if(ref_ptr == null)
{
head = head.next;
if(head != null)
Console.WriteLine("Node no. " +
n + " from last is " +
main_ptr.data);
}
else
{
while (ref_ptr != null)
{
main_ptr = main_ptr.next;
ref_ptr = ref_ptr.next;
}
Console.WriteLine("Node no. " +
n + " from last is " +
main_ptr.data);
}
}
}
/* 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;
}
/*Driver code */
public static void Main(String[] args)
{
LinkedList llist = new LinkedList();
llist.push(20);
llist.push(4);
llist.push(15);
llist.push(35);
llist.printNthFromLast(4);
}
}
/* This code is contributed by PrinciRaj1992 */
|
Javascript
Output
10->20->30->40->50->60->NULL
Node no. 1 from end is: 60
Node no. 2 from end is: 50
Node no. 3 from end is: 40
Node no. 4 from end is: 30
Node no. 5 from end is: 20
Node no. 6 from end is: 10
7 is greater than no. of nodes in the list
Time Complexity: O(n) where n is the length of linked list.
Please write comments if you find the above codes/algorithms incorrect, or find other ways to solve the same problem.
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Linked List
C++
// C++ program to find n'th // node in linked list #include #include using namespace std; // Link list node class Node { public: int data; Node* next; }; /* Given a reference (pointer to pointer) to the head of a list and an int, push a new node on the front of the list. */ void 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 // off the new node new_node->next = (*head_ref); // move the head to point // to the new node (*head_ref) = new_node; } // Takes head pointer of // the linked list and index // as arguments and return // data at index int GetNth(Node* head,int index) { Node* current = head; // the index of the // node we're currently // looking at int count = 0; while (current != NULL) { if (count == index) return(current->data); count++; current = current->next; } /* if we get to this line, the caller was asking for a non-existent element so we assert fail */ assert(0); } // Driver Code int main() { // Start with the // empty list Node* head = NULL; // Use push() to construct // below list // 1->12->1->4->1 push(&head, 1); push(&head, 4); push(&head, 1); push(&head, 12); push(&head, 1); // Check the count // function cout <<"Element at index 3 is " << GetNth(head, 3); return 0; } // This code is contributed by rathbhupendra |
Write a function to get Nth node in a Linked List in C++
C++Server Side ProgrammingProgramming
Here, we are given a linked list and an index. We have to write a function to get Nth node in a linked list.
Let’s take an example to understand the problem,
Iterative Solution
A simple solution is to calculate the total number of nodes n in the linked list first. Then, the k'th node from the end will be (n-k+1)'th node from the beginning.
Following is the C, Java, and Python program that demonstrates it:
