Find the middle element in a linked list

In this article, we will discuss how to find the middle element in a linked list. A linked list has several nodes, and each node in the list has the content and a pointer to the next node in the list. It does not store any pointer to the previous node. To store a single linked list, only the pointer to the first node in that list must be stored. The last node in a single linked list points to nothing.

In our previous article, we have discussed how to find the nth element in a singly linked list. This problem almost the same as the previous problem.

Find the middle element in a linked list

This is one of a very frequently asked question of the interview questions. There are several ways to explain the answer to this question. But interviewers ask for the efficient way to find the middle element in a linked list. Let’s see the data structure class of the linked list like the following:

package com.dineshonjava.algo;

/**
 * @author Dinesh.Rajput
 *
 */
public class Node {
	
	private Node next;
    private String data;
    
    public Node(String data) {
		super();
		this.data = data;
	}

	public boolean hasNext() {
        return next != null;
    }
 
    public void setNext(Node next) {
        this.next = next;
    }
    
    public Node getNext() {
		return next;
	}
    
	public String getData() {
		return data;
	}

	public String toString() {
        return this.data;
    }
}

Let’s see the following answers with several assumptions.

Assumption 1: Using Size of the Linked List

Step 1: First, traverse the whole linked list and find the size of the linked list.

Step 2: After finding a size of the linked again traverse to size/2 and locate size/2 element from the head of linkedlist.

Let’s see the following example:

/**
 * Find the middle element in a linked list using size of the linked list
 */
package com.dineshonjava.algo;

import java.util.Optional;

/**
 * @author Dinesh.Rajput
 *
 */
public class LinkedListTest {

	/**
	 * @param args
	 */
	public static void main(String[] args) {
		
		Node listNode = createLinkedList(11);
		System.out.println(findMiddleElementInLinkedList(listNode).get());
	}
	
	private static Node createLinkedList(int n) {
	    Node head = new Node("1");
	    Node current = head;
	 
	    for (int i = 2; i <= n; i++) {
	        Node newNode = new Node(String.valueOf(i));
	        current.setNext(newNode);
	        current = newNode;
	    }
	 
	    return head;
	}
	
	public static Optional findMiddleElementInLinkedList(Node head) {
	    if (head == null) {
	        return Optional.empty();
	    }
	 
	    Node current = head;
	    int size = 1;
	    while (current.hasNext()) {
	        current = current.getNext();
	        size++;
	    }
	 
	    current = head;
	    for (int i = 0; i < (size - 1) / 2; i++) {
	        current = current.getNext();
	    }
	 
	    return Optional.of(current.getData());
	}
}

You can run this code and find the middle element in a linked list using the size of the linked list. But in this approach the time complexity = time for finding the length of the list + time for locating the middle element. That means, total time complexity = o(n) + o(n) = o(n) and here the space complexity= o(1).

Let’s discuss another efficient way to find the middle element in a linked list without using the size of the linked list.

Assumptions 2: Without using Size of the Linked List

In this approach, we will traverse this linked list with the two pointers.

Step 1: Let’s assume two pointers as slowPointer and fastPointer and initialize them with the head.

Step 2: And slowPointer moves one by one node but teh fastPointer moves faster by two nodes.

Step 3: When the fastPointer will reach the end of the linked list then slowPoint will be at the middle node, is it point we are looking for.

Let’s see the code like the following:

/**
 * Find the middle element in a linked list without using size of the linked list
 */
package com.dineshonjava.algo;

import java.util.Optional;

/**
 * @author Dinesh.Rajput
 *
 */
public class LinkedListTest {

	/**
	 * @param args
	 */
	public static void main(String[] args) {
		
		Node listNode = createLinkedList(11);
		System.out.println(findMiddleElementInLinkedList(listNode).get());
	}
	
	private static Node createLinkedList(int n) {
	    Node head = new Node("1");
	    Node current = head;
	 
	    for (int i = 2; i <= n; i++) {
	        Node newNode = new Node(String.valueOf(i));
	        current.setNext(newNode);
	        current = newNode;
	    }
	 
	    return head;
	}
	
	private static Optional findMiddleElementInLinkedList(Node head) {
		if (head == null) {
	        return Optional.empty();
	    }
	 
	    Node slowPointer = head;
	    Node fastPointer = head;
	 
	    while (fastPointer.hasNext() && fastPointer.getNext().hasNext()) {
	        fastPointer = fastPointer.getNext().getNext();
	        slowPointer = slowPointer.getNext();
	    }
	    return Optional.ofNullable(slowPointer.getData());
	}
	
	

In this example, we can find the middle element in a linked list in a single traversal.

Hope, you have understood this solution for the above find the middle node from a linked list. Please share other solutions if you have. :).

Happy learning with us!!!.