Write a function to get Nth node in a Linked List
Last Updated : 23 Aug, 2024
Given a LinkedList and an index (1-based). The task is to find the data value stored in the node at that kth position. If no such node exists whose index is k then return -1.
Example:
Input: 1->10->30->14, index = 2
Output: 10
Explanation: The node value at index 2 is 10
Input: 1->32->12->10->30->14->100, index = 8
Output: -1
Explanation: No such node exists at index = 8.
[Naive Approach] Recursive Method - O(n) Time and O(n) Space
The idea is to use the recursive method to find the value of index node (1- based) . Call the function GetNth(head,index) recusively, where head will represent the current head node . Decrement the index value by 1 on every recursion call. When the n reaches 1 ,we will return the data of current node.
Below is the implementation of above approach:
C++ //C++ program to find the data at nth node //recursively #include <bits/stdc++.h> using namespace std; struct Node { int data; Node* next; Node(int x) { data = x; next = NULL; } }; // Takes head pointer of the linked list and index // as arguments and returns data at index. int GetNth(Node* head, int index) { // If the list is empty or index is out of bounds if (head == NULL) return -1; // If index equals 1, return node's data if (index == 1) return head->data; // Recursively move to the next node return GetNth(head->next, index - 1); } int main() { // Create a hard-coded linked list: // 1 -> 2 -> 3 -> 4 -> 5 Node* head = new Node(1); head->next = new Node(2); head->next->next = new Node(3); head->next->next->next = new Node(4); head->next->next->next->next = new Node(5); cout << "Element at index 3 is " << GetNth(head, 3) << endl; return 0; } // C program to find the data at nth node // recursively #include <stdio.h> struct Node { int data; struct Node *next; }; // Takes head pointer of the linked list and index // as arguments and returns data at index. int GetNth(struct Node *head, int index) { // If the list is empty or index is out of bounds if (head == NULL) return -1; // If index equals 1, return node's data if (index == 1) return head->data; // Recursively move to the next node return GetNth(head->next, index - 1); } struct Node *createNode(int new_data) { struct Node *new_node = (struct Node *)malloc(sizeof(struct Node)); new_node->data = new_data; new_node->next = NULL; return new_node; } int main() { // Create a hard-coded linked list: // 1 -> 2 -> 3 -> 4 -> 5 struct Node *head = createNode(1); head->next = createNode(2); head->next->next = createNode(3); head->next->next->next = createNode(4); head->next->next->next->next = createNode(5); printf("Element at index 3 is %d\n", GetNth(head, 3)); return 0; } // Java program to find n'th node in linked list // using recursion import java.io.*; class Node { int data; Node next; Node(int x){ data = x; next = null; } } class GfG { // Takes head pointer of the linked list and index // as arguments and return data at index*/ static int GetNth(Node head, int index) { if (head == null) return -1; // if index equal to 1 return node.data if (index == 1) return head.data; // recursively decrease n and increase // head to next pointer return GetNth(head.next, index - 1); } public static void main(String args[]) { // Create a hard-coded linked list: // 1 -> 2 -> 3 -> 4 -> 5 Node head = new Node(1); head.next = new Node(2); head.next.next = new Node(3); head.next.next.next = new Node(4); head.next.next.next.next = new Node(5); System.out.printf("Element at index 3 is %d", GetNth(head, 3)); } } # Python program to find the Nth node in # linked list using recursion class Node: def __init__(self, x): self.data = x self.next = None # Recursive method to find the Nth node def get_nth_node(head, index): # Helper function to handle recursion #and count tracking if head is None: print(-1) if index == 1: print(head.data) else: get_nth_node(head.next, index-1) if __name__ == "__main__": # Create a linked list: 1 -> 2 -> 3 -> 4 -> 5 head = Node(1) head.next = Node(2) head.next.next = Node(3) head.next.next.next = Node(4) head.next.next.next.next = Node(5) print("Element at index 3 is", end=" ") get_nth_node(head, 3) // C# program to find the Nth node in // linked list using recursion using System; class Node { public int Data; public Node Next; public Node(int x) { Data = x; Next = null; } } class GfG { // Takes head pointer of the linked list and index // as arguments and returns data at index static int GetNth(Node head, int index) { // Base Condition if (head == null) return -1; // If n equals 0, return the node's data if (index == 1) return head.Data; // Recursively move to the next node return GetNth(head.Next, index - 1); } public static void Main() { // Create a hard-coded linked list: // 1 -> 2 -> 3 -> 4 -> 5 Node head = new Node(1); head.Next = new Node(2); head.Next.Next = new Node(3); head.Next.Next.Next = new Node(4); head.Next.Next.Next.Next = new Node(5); Console.WriteLine("Element at index 3 is {0}", GetNth(head, 3)); } } // JavaScript program to find the n'th node in // a linked list using recursion class Node { constructor(new_data) { this.data = new_data; this.next = null; } } function GetNth(head, index) { // Base case: if the list is empty or index is out of // bounds if (head === null) { return -1; } // Base case: if count equals n, return node's data if (index === 1) { return head.data; } // Recursive case: move to the next node and decrease // index return GetNth(head.next, index - 1); } // Create a hard-coded linked list: // 1 -> 2 -> 3 -> 4 -> 5 let head = new Node(1); head.next = new Node(2); head.next.next = new Node(3); head.next.next.next = new Node(4); head.next.next.next.next = new Node(5); console.log("Element at index 3 is", GetNth(head, 3)); OutputElement at index 3 is 3
Time Complexity : O(n) ,where n is the nth node of linked list.
Auxiliary Space: O(n), for recursive call stack
[Expected Approach-2] Iterative Method - O(n) Time and O(1) Space
The idea is similar to recursive approach to find the value at index node (1- based) .We will use a variable say, count = 1 to track the nodes. Traverse the list until curr != NULL . Increment the count if count is not equal to index node (1- based) , else if count equals to the index node, return data at current node.
Below is the implementation of above approach :
C++ // C++ program to find n'th // node in linked list (iteratively) #include <iostream> using namespace std; class Node { public: int data; Node *next; Node(int x) { data = x; next = nullptr; } }; // Function to find the nth node in the list int GetNth(Node *head, int index) { Node *curr = head; int count = 1; while (curr != nullptr) { if (count == index) return curr->data; count++; curr = curr->next; } return -1; } int main() { // Create a hard-coded linked list: // 1 -> 2 -> 3 -> 4 -> 5 Node *head = new Node(1); head->next = new Node(2); head->next->next = new Node(3); head->next->next->next = new Node(4); head->next->next->next->next = new Node(5); cout << "Element at index 3 is " << GetNth(head, 3) << endl; return 0; } // C program to find n'th // node in linked list (iteratively) #include <stdio.h> struct Node { int data; struct Node *next; }; // Function to find the nth node in the list int GetNth(struct Node *head, int index) { struct Node *curr = head; int count = 1; while (curr != NULL) { if (count == index) return curr->data; count++; curr = curr->next; } return -1; } struct Node *createNode(int new_data) { struct Node *new_node = (struct Node *)malloc(sizeof(struct Node)); new_node->data = new_data; new_node->next = NULL; return new_node; } int main() { // Create a hard-coded linked list: // 1 -> 2 -> 3 -> 4 -> 5 struct Node *head = createNode(1); head->next = createNode(2); head->next->next = createNode(3); head->next->next->next = createNode(4); head->next->next->next->next = createNode(5); printf("Element at index 3 is %d\n", GetNth(head, 3)); } // Java program to find the Nth node in // a linked list iteratively class Node { int data; Node next; Node(int x) { data = x; next = null; } } class GfG { // Function to find the nth node in the list iteratively static int getNthNodeIterative(Node head, int index) { Node current = head; int count = 1; // Traverse the list until the end or until the nth // node is reached while (current != null) { if (count == index) { return current.data; } count++; current = current.next; } // Return -1 if the index is out of bounds return -1; } public static void main(String[] args) { // Create a hard-coded linked list: // 1 -> 2 -> 3 -> 4 -> 5 Node head = new Node(1); head.next = new Node(2); head.next.next = new Node(3); head.next.next.next = new Node(4); head.next.next.next.next = new Node(5); int index = 3; int result = getNthNodeIterative(head, index); if (result != -1) { System.out.println("Element at index " + index + " is " + result); } else { System.out.println("Index " + index + " is out of bounds"); } } } # Python program to find the Nth node in # a linked list iteratively class Node: def __init__(self, x): self.data = x self.next = None # Function to find the nth node in the list iteratively def get_nth_node_iterative(head, n): current = head count = 1 # Traverse the list until the end or until the nth node is reached while current is not None: if count == n: return current.data count += 1 current = current.next # Return -1 if the index is out of bounds return -1 if __name__ == "__main__": # Create a hard-coded linked list: # 1 -> 2 -> 3 -> 4 -> 5 head = Node(1) head.next = Node(2) head.next.next = Node(3) head.next.next.next = Node(4) head.next.next.next.next = Node(5) index = 3 result = get_nth_node_iterative(head, index) if result != -1: print(f"Element at index {index} is {result}") else: print(f"Index {index} is out of bounds") // Iterative C# program to find the nth node in // a linked list using System; class Node { public int Data; public Node Next; public Node(int x) { Data = x; Next = null; } } class GfG { // Given the head of a list and index, find the nth node // and return its data static int GetNthNode(Node head, int n) { Node current = head; int count = 1; // Traverse the list until the nth node is found or // end of the list is reached while (current != null) { if (count == n) { return current.Data; } count++; current = current.Next; } // Return -1 if the index is out of bounds return -1; } public static void Main() { // Create a hard-coded linked list: // 1 -> 2 -> 3 -> 4 -> 5 Node head = new Node(1); head.Next = new Node(2); head.Next.Next = new Node(3); head.Next.Next.Next = new Node(4); head.Next.Next.Next.Next = new Node(5); int index = 3; int result = GetNthNode(head, index); if (result != -1) { Console.WriteLine($"Element at index {index} is {result}"); } else { Console.WriteLine($"Index {index} is out of bounds"); } } } // Iterative JavaScript program to find the Nth node in a // linked list class Node { constructor(x) { this.data = x; this.next = null; } } // Given the head of a list and an index, return the data at // the index function getNth(head, index) { let current = head; let count = 1; // Traverse the linked list while (current !== null) { if (count === index) { // Return data at the current // node if index matches return current.data; } count++; current = current.next; } // Return -1 if index is out of bounds return -1; } // Create a hard-coded linked list: // 1 -> 2 -> 3 -> 4 -> 5 let head = new Node(1); head.next = new Node(2); head.next.next = new Node(3); head.next.next.next = new Node(4); head.next.next.next.next = new Node(5); let index = 3; let result = getNth(head, index); if (result !== -1) { console.log(`Element at index ${index} is ${result}`); } else { console.log(`Index ${index} is out of bounds`); } OutputElement at index 3 is 3
Time Complexity : O(n), where n is the nth node of linked list.
Auxiliary Space: O(1)
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