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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Python3
# Python3 program for Recursive approach # to find middle of singly linked list # Node class class Node: # Function to initialise the node object def __init__(self, data): self.data = data self.next = None # Create new Node def newLNode(data): temp = Node(data) temp.data = data temp.next = None return temp mid = Nonen = 0 # Function for finding midpoint recursively def midpoint_util(head ): global n global mid # If we reached end of linked list if (head == None): n = int((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 def midpoint(head): global n global mid mid = None n = 1 midpoint_util(head) return mid # Driver Code if __name__=='__main__': 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) result = midpoint(head) 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 |
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Output:
3
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