Find middle of singly linked list Recursively
Given a singly linked list and the task is to find the middle of the linked list.
Examples:
Input : 1->2->3->4->5 Output : 3 Input : 1->2->3->4->5->6 Output : 4
We have already discussed Iterative Solution. In this post iterative solution is discussed. Count total number of nodes in the list in recursive manner and do half of this, suppose this value is n. Then rolling back through recursion decrement n by one for each call. Return the node where n is zero.
C++
// C++ program for Recursive approach to find // middle of singly linked list #include <iostream> using namespace std; // Tree Node Structure struct Node { int data; struct Node* next; }; // Create new Node Node* newLNode(int data) { Node* temp = new Node; temp->data = data; temp->next = NULL; return temp; } // Function for finding midpoint recursively void midpoint_util(Node* head, int* n, Node** mid) { // If we reached end of linked list if (head == NULL) { *n = (*n) / 2; return; } *n = *n + 1; midpoint_util(head->next, n, mid); // Rolling back, decrement n by one *n = *n - 1; if (*n == 0) { // Final answer *mid = head; } } Node* midpoint(Node* head) { Node* mid = NULL; int n = 1; midpoint_util(head, &n, &mid); return mid; } int main() { Node* head = newLNode(1); head->next = newLNode(2); head->next->next = newLNode(3); head->next->next->next = newLNode(4); head->next->next->next->next = newLNode(5); Node* result = midpoint(head); cout << result->data << endl; return 0; } |
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Java
// Java program for Recursive approach to find // middle of singly linked list class GFG { // Tree Node Structure static class Node { int data; Node next; }; // Create new Node static Node newLNode(int data) { Node temp = new Node(); temp.data = data; temp.next = null; return temp; } static int n; static Node mid; // Function for finding midpoint recursively static void midpoint_util(Node head ) { // If we reached end of linked list if (head == null) { n = (n) / 2; return; } n = n + 1; midpoint_util(head.next); // Rolling back, decrement n by one n = n - 1; if (n == 0) { // Final answer mid = head; } } static Node midpoint(Node head) { mid = null; n = 1; midpoint_util(head); return mid; } // Driver code public static void main(String args[]) { Node head = newLNode(1); head.next = newLNode(2); head.next.next = newLNode(3); head.next.next.next = newLNode(4); head.next.next.next.next = newLNode(5); Node result = midpoint(head); System.out.print( result.data ); } } // This code is contributed by Arnab Kundu |
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C#
// C# program for Recursive approach to find
// middle of singly linked list
using System;
class GFG
{
// Tree Node Structure
public class Node
{
public int data;
public Node next;
};
// Create new Node
static Node newLNode(int data)
{
Node temp = new Node();
temp.data = data;
temp.next = null;
return temp;
}
static int n;
static Node mid;
// Function for finding midpoint recursively
static void midpoint_util(Node head )
{
// If we reached end of linked list
if (head == null)
{
n = (n) / 2;
return;
}
n = n + 1;
midpoint_util(head.next);
// Rolling back, decrement n by one
n = n – 1;
if (n == 0)
{
// Final answer
mid = head;
}
}
static Node midpoint(Node head)
{
mid = null;
n = 1;
midpoint_util(head);
return mid;
}
// Driver code
public static void Main()
{
Node head = newLNode(1);
head.next = newLNode(2);
head.next.next = newLNode(3);
head.next.next.next = newLNode(4);
head.next.next.next.next = newLNode(5);
Node result = midpoint(head);
Console.WriteLine( result.data );
}
}
// This code is contributed by Rajput-Ji
Output:
3
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