Intersection of Two Linked Lists

Write a program to find the node at which the intersection of two singly linked lists begins.

For example, the following two linked lists: 

A:          a1 → a2
                   ↘
                     c1 → c2 → c3
                   ↗            
B:     b1 → b2 → b3

begin to intersect at node c1.

Notes:

• If the two linked lists have no intersection at all, return null.
• The linked lists must retain their original structure after the function returns. 
• You may assume there are no cycles anywhere in the entire linked structure.
• Your code should preferably run in O(n) time and use only O(1) memory.

A fresh new problem. Somehow I am very in to linked list. I think I am just having fun play around with pointers. 

Back to the problem. I found this method through GeeksforGeeks, and implemented it using Java. Consider I am such a fake one, I have no guilty borrowing it. 

1. Determine the size of both linked lists.  

2. Get the difference between two lists diff = abs(sizeA - sizeB). 

3. Traverse the list with the larger size until node diff. 

4. Traverse both lists together, if both the value and pointer of the nodes are the same in two lists, we can conclude they intersect. Otherwise, if nothing pops up till the end of both lists, return null. 

Mission accomplished. :)

public class ListNode {
      int val;
      ListNode next;
      ListNode(int x) {
          val = x;
          next = null;
      }
  }
 
public class Solution {
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        if (headA == null || headB == null)
            return null;
        int sizeA = getSize(headA);
        int sizeB = getSize(headB);
        int diff = Math.abs(sizeA - sizeB);
        int index = 0;
        if (sizeA > sizeB)
        {
            while (headA != null && index < diff)
            {
                headA = headA.next;
                index++;
            }
        }
        else
        {
            while(headB != null && index < diff)
            {
                headB = headB.next;
                index++;
            }
        }
        while (headA != null && headB != null)
        {
            if (headA.val == headB.val && headA.next == headB.next)
                return headA;
            headA = headA.next;
            headB = headB.next;
        }
        return null;
        
        
    }
    private int getSize(ListNode head)
    {
        int size = 0;
        while (head != null)
        {
            size++;
            head = head.next;
        }
        return size;
    }
}