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”
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 <bits/stdc++.h>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 Codeint 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 listclass 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 endclass Node: def __init__(self, new_data): self.data = new_data self.next = None class LinkedList: def __init__(self): self.head = None # createNode and 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
<script>// Simple Javascript program to find n'th node from end of linked list /* Linked List node */ class Node { constructor(d) { this.data = d; this.next = null; } } /* Function to get the nth node from the last of a linked list */ class LinkedList { constructor(d){ this.head = d; } printNthFromLast(n) { let len = 0; let temp = this.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 = this.head; // 2) get the (len-n+1)th node from the beginning for (let i = 1; i < len - n + 1; i++) temp = temp.next; document.write(temp.data); } /* Inserts a new Node at front of the list. */ push(new_data) { /* 1 & 2: Allocate the Node & Put in the data*/ let new_node = new Node(new_data); /* 3. Make next of new Node as head */ new_node.next = this.head; /* 4. Move the head to point to new Node */ this.head = new_node; } } /*Driver program to test above methods */ let llist = new LinkedList(); llist.push(20); llist.push(4); llist.push(15); llist.push(35); llist.printNthFromLast(4); // This code is contributed by Saurabh Jaiswal</script> |
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<<head->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
<script>function printNthFromLast(head , n){ function i = 0; if (head == null) return; printNthFromLast(head.next, n); if (++i == n) document.write(head.data);}// This code is contributed by gauravrajput1</script> |
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 <bits/stdc++.h>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 pointersclass 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 = Noneclass 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
<script>// javascript program to find n'th // node from end using slow and// fast pointersvar head; // head of the list /* Linked List node */ class Node { constructor(val) { this.data = val; this.next = null; } } /* * Function to get the nth node from end of list */ function printNthFromLast(n) { var main_ptr = head; var ref_ptr = head; var count = 0; if (head != null) { while (count < n) { if (ref_ptr == null) { document.write(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) document.write("Node no. " + n + " from last is " + head.data); } else { while (ref_ptr != null) { main_ptr = main_ptr.next; ref_ptr = ref_ptr.next; } document.write("Node no. " + n + " from last is " + main_ptr.data); } } } /* Inserts a new Node at front of the list. */ function push(new_data) { /* * 1 & 2: Allocate the Node & Put in the data */ var 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 */ push(20); push(4); push(15); push(35); printNthFromLast(4);// This code is contributed by Rajput-Ji </script> |
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.
