CareerCup

using two pointers

p1=head;
p2=head->next;
while(p2!=null) 
{   
    p1=p1->next;
    p2=p2->next
    if(p2!=null) p2=p2->next;
}
return p1;

Logic is - increment first pointer, and increment second pointer twice as fast. When second pointer reaches the end, the first one will point to the middle ( because it was moving 2 times slower).

typedef struct Node {
int data;
Node* next;
};

Node* getMiddle(Node* head) {
int count = 0;
Node* middle = head;
Node* curr = head;
while (curr != NULL) {
if (count == 2) {
middle = middle->next;
count = 0;
}
curr = curr->next;
count++;
}

return middle;
}

middle=head;
last=head;
int count = 0;
while(last)
{
last = last->next;
if(count % 2)
{
middle = middle->next
}
++count;
}
return middle;

What kind of Singly linked list are you referring to? Please clarify ...

1. Basic singly linkedlist
2. Double "ended" sinlgly linked list (pointer to the head and the tail elements of the list
3. Circular singly linkedlist
4. Singly linked list w/ a sentinel

Can someone explain the logic? or preferable approach in Java

Wow...! That's the best solution.

Here is the code for java programmers

static void findMiddleElement(LinkedListNode head) {
		// LinkedListNode head = new LinkedListNode(1);
		// head.appendToTail(2);
		// head.appendToTail(3);
		// head.appendToTail(4);
		// head.appendToTail(5);
		// head.appendToTail(6);
		// head.appendToTail(7);
		// head.appendToTail(8);
		// head.appendToTail(9);
		// head.appendToTail(10);
		// head.appendToTail(11);
		
		LinkedListNode current = head;
		LinkedListNode middle = head;

		while (current.next != null) {
			current = current.next;
			if (current.next != null) {
				current = current.next;
				middle = middle.next;
			}
		}
		System.out.println("Middle Element :" + middle.value);
	}


public class LinkedListNode {
	public LinkedListNode next;
	public LinkedListNode previous;
	public int value;

	public LinkedListNode(int value) {
		this.value = value;
	}

	public void appendToTail(int value) {
		LinkedListNode end = new LinkedListNode(value);
		LinkedListNode temp = this;
		while (temp.next != null) {
			temp = temp.next;
		}
		temp.next = end;
	}}

def solution_9