Finding middle element in a linked list (DSA)

 

Problem Statement: (👈click here for GFG)

Given a singly linked list of N nodes.
The task is to find the middle of the linked list. For example, if the linked list is
1-> 2->3->4->5, then the middle node of the list is 3.
If there are two middle nodes(in case, when N is even), print the second middle element.
For example, if the linked list given is 1->2->3->4->5->6, then the middle node of the list is 4.

Example 1:

Input:
LinkedList: 1->2->3->4->5
Output: 3 
Explanation: 
Middle of linked list is 3.

Example 2: 

Input:
LinkedList: 2->4->6->7->5->1
Output: 7 
Explanation: 
Middle of linked list is 7.

Your Task:
The task is to complete the function getMiddle() which takes a head reference as the only argument and should return the data at the middle node of the linked list.

Expected Time Complexity: O(N).
Expected Auxiliary Space: O(1).

Constraints:
1 <= N <= 5000

Solution:

class Solution {
    /* Should return data of middle node. If linked list is empty, then  -1*/
    getMiddle(node)
    {
        let slow = node;
        let fast = node;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        return slow.data;
    }
}