Given a Linked List and a number K. The task is to print the value of the K-th node from the middle towards the beginning of the List. If no such element exists, then print “-1”.
Note: Position of middle node is: (n/2)+1, where n is the total number of nodes in the list.
Examples:
Input : List is 1->2->3->4->5->6->7 K= 2 Output : 2 Input : list is 7->8->9->10->11->12 K = 3 Output : 7
Traverse the List from beginning to end and count the total number of nodes. Now, suppose is the total number of nodes in the List. Therefore, the middle node will be at the position (n/2)+1. Now, the task remains to print the node at (n/2 + 1 – k)th position from the head of the List.
Below is the implementation of the above idea:
C++
// CPP program to find kth node from middle // towards Head of the Linked List #include <bits/stdc++.h> using namespace std; // Linked list node struct Node { int data; struct 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( struct Node** head_ref, int new_data) { struct Node* new_node = new Node; new_node->data = new_data; new_node->next = (*head_ref); (*head_ref) = new_node; } // Function to count number of nodes int getCount( struct Node* head) { int count = 0; // Initialize count struct Node* current = head; // Initialize current while (current != NULL) { count++; current = current->next; } return count; } // Function to get the kth node from the mid // towards begin of the linked list int printKthfrommid( struct Node* head_ref, int k) { // Get the count of total number of // nodes in the linked list int n = getCount(head_ref); int reqNode = ((n / 2 + 1) - k); // If no such node exists, return -1 if (reqNode <= 0) { return -1; } // Find node at position reqNode else { struct Node* current = head_ref; // the index of the // node we're currently // looking at int count = 1; while (current != NULL) { if (count == reqNode) return (current->data); count++; current = current->next; } } } // Driver code int main() { // start with empty list struct Node* head = NULL; int k = 2; // create linked list // 1->2->3->4->5->6->7 push(&head, 7); push(&head, 6); push(&head, 5); push(&head, 4); push(&head, 3); push(&head, 2); push(&head, 1); cout << printKthfrommid(head, 2); return 0; } |
Java
// Java program to find Kth node from mid // of a linked list in single traversal // Linked list node class Node { int data; Node next; Node( int d) { this .data = d; this .next = null ; } } class LinkedList { Node start; LinkedList() { start = null ; } // Function to push node at head public void push( int data) { if ( this .start == null ) { Node temp = new Node(data); this .start = temp; } else { Node temp = new Node(data); temp.next = this .start; this .start = temp; } } //method to get the count of node public int getCount(Node start) { Node temp = start; int cnt = 0 ; while (temp != null ) { temp = temp.next; cnt++; } return cnt; } // Function to get the kth node from the mid // towards begin of the linked list public int printKthfromid(Node start, int k) { // Get the count of total number of // nodes in the linked list int n = getCount(start); int reqNode = ((n + 1 ) / 2 ) - k; // If no such node exists, return -1 if (reqNode <= 0 ) return - 1 ; else { Node current = start; int count = 1 ,ans = 0 ; while (current != null ) { if (count == reqNode) { ans = current.data; break ; } count++; current = current.next; } return ans; } } // Driver code public static void main(String[] args) { // create linked list // 1->2->3->4->5->6->7 LinkedList ll = new LinkedList(); ll.push( 7 ); ll.push( 6 ); ll.push( 5 ); ll.push( 4 ); ll.push( 3 ); ll.push( 2 ); ll.push( 1 ); System.out.println(ll.printKthfromid(ll.start, 2 )); } } // This Code is contributed by Adarsh_Verma |
Python3
# Python3 program to find kth node from # middle towards Head of the Linked List # Node class class Node: # Function to initialise the node object def __init__( self , data): self .data = data self . next = None # Function to insert a node at the # beginning of the linked list def push(head, data): if not head: # Return new node return Node(data) # allocate node new_node = Node(data) # link the old list to the new node new_node. next = head # move the head to point # to the new node head = new_node return head # Function to count number of nodes def getCount(head): count = 0 # Initialize count current = head # Initialize current while (current ): count = count + 1 current = current. next return count # Function to get the kth node from the mid # towards begin of the linked list def printKthfrommid(head_ref, k): # Get the count of total number of # nodes in the linked list n = getCount(head_ref) reqNode = int ((n / 2 + 1 ) - k) # If no such node exists, return -1 if (reqNode < = 0 ) : return - 1 # Find node at position reqNode else : current = head_ref # the index of the # node we're currently # looking at count = 1 while (current) : if (count = = reqNode): return (current.data) count = count + 1 current = current. next # Driver Code if __name__ = = '__main__' : # start with empty list head = None k = 2 # create linked list # 1.2.3.4.5.6.7 head = push(head, 7 ) head = push(head, 6 ) head = push(head, 5 ) head = push(head, 4 ) head = push(head, 3 ) head = push(head, 2 ) head = push(head, 1 ) print (printKthfrommid(head, 2 )) # This code is contributed by Arnab Kundu |
C#
// C# program to find Kth node from mid // of a linked list in single traversal using System; // Linked list node public class Node { public int data; public Node next; public Node( int d) { this .data = d; this .next = null ; } } public class LinkedList { Node start; LinkedList() { start = null ; } // Function to push node at head public void push( int data) { if ( this .start == null ) { Node temp = new Node(data); this .start = temp; } else { Node temp = new Node(data); temp.next = this .start; this .start = temp; } } //method to get the count of node public int getCount(Node start) { Node temp = start; int cnt = 0; while (temp != null ) { temp = temp.next; cnt++; } return cnt; } // Function to get the kth node from the mid // towards begin of the linked list public int printKthfromid(Node start, int k) { // Get the count of total number of // nodes in the linked list int n = getCount(start); int reqNode = ((n + 1) / 2) - k; // If no such node exists, return -1 if (reqNode <= 0) return -1; else { Node current = start; int count = 1,ans = 0; while (current != null ) { if (count == reqNode) { ans = current.data; break ; } count++; current = current.next; } return ans; } } // Driver code public static void Main(String[] args) { // create linked list // 1->2->3->4->5->6->7 LinkedList ll = new LinkedList(); ll.push(7); ll.push(6); ll.push(5); ll.push(4); ll.push(3); ll.push(2); ll.push(1); Console.WriteLine(ll.printKthfromid(ll.start, 2)); } } // This code is contributed by Rajput-Ji |
2
Time Complexity: O(n), where n is the length of the list.
Auxiliary Space: O(1)
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