Following code is failing to find intersection of two lists

Question

I am doing LC:60 which is about finding intersection of two linked lists. I wrote following code based on this logic

• find length of both lists.
• advance 1 pointer of long list to the difference of the length
• compare both lists from this point onwards.

This is working for all test-cases except for test case where two lists have same input i.e., [2, 2, 4, 5, 4] but the intersection node is node 4.

Here is the code

public class Solution {
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        int aN = 0;
        ListNode a = headA;
        while (a != null) {
            aN++;
            a = a.next;
        }    

        int bN = 0;
        ListNode b = headB;
        while (b != null) {
            bN++;
            b = b.next;
        }

        ListNode longList = (aN > bN) ? headA : headB;
        ListNode shortList = (aN < bN) ? headA : headB;

        for (int i = 0; i < Math.abs(aN - bN); i++) {
            longList = longList.next;
        }

        while (longList != null && shortList != null) {
            if (longList == shortList) { return longList; }
            longList = longList.next;
            shortList = shortList.next;
        }

        return null;
    }
}

Answer 1

The problem is in these statements:

    ListNode longList = (aN > bN) ? headA : headB;
    ListNode shortList = (aN < bN) ? headA : headB;

If both lists have the same size, then both longList and shortList will be set to headB. You should have a ternary condition that is the exact opposite (<= for the second one), or maybe more intuitive, use the same condition but swap the operands in the second assignment:

    ListNode longList = (aN > bN) ? headA : headB;
    ListNode shortList = (aN > bN) ? headB : headA;