Auxiliary Given a singly linked list, find the middle of the linked list. For example, if the given linked list is 1->2->3->4->5 then the output should be 3. If there are even nodes, then there would be two middle nodes, we need to print the second middle element. For example, if the given linked list is 1->2->3->4->5->6 then the output should be 4.

To solve this problem, we can use the two-pointer technique. Here are the steps:

1. Initialize two pointers, slow_ptr and fast_ptr, to head of the linked list.
2. Move slow_ptr by one and fast_ptr by two.
3. When fast_ptr reaches the end of the linked list, slow_ptr will be at the middle of the linked list.

Here is the Python code for the above steps:

class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

def findMiddle(head):
    slow_ptr = head
    fast_ptr = head

    if head is not None:
        while(fast_ptr is not None and fast_ptr.next is not None):
            fast_ptr = fast_ptr.next.next
            slow_ptr = slow_ptr.next

    return slow_ptr.data

In the above code, we first define a Node class to represent each node in the linked list. Then we define a function findMiddle that takes the head of the linked list as input and returns the data of the middle node. Inside the function, we initialize slow_ptr and fast_ptr to head. Then we enter a loop that continues until fast_ptr reaches the end of the linked list. Inside the loop, we move fast_ptr by two and slow_ptr by one. When the loop ends, slow_ptr will be at the middle of the linked list, so we return its data.