/**
 * Write a program to find the node at which the intersection of two singly linked lists begins.
 *
 * For example, the following two linked lists:
 *
 *
 * begin to intersect at node c1.
 *
 * Example 1:
 *
 * Input: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,0,1,8,4,5], skipA = 2, skipB = 3
 * Output: Reference of the node with value = 8
 * Input Explanation: The intersected node's value is 8 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [4,1,8,4,5]. From the head of B, it reads as [5,0,1,8,4,5]. There are 2 nodes before the intersected node in A; There are 3 nodes before the intersected node in B.
 *
 *
 * Example 2:
 *
 * Input: intersectVal = 2, listA = [0,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1
 * Output: Reference of the node with value = 2
 * Input Explanation: The intersected node's value is 2 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [0,9,1,2,4]. From the head of B, it reads as [3,2,4]. There are 3 nodes before the intersected node in A; There are 1 node before the intersected node in B.
 *
 *
 * Example 3:
 *
 * Input: intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2
 * Output: null
 * Input Explanation: From the head of A, it reads as [2,6,4]. From the head of B, it reads as [1,5]. Since the two lists do not intersect, intersectVal must be 0, while skipA and skipB can be arbitrary values.
 * Explanation: The two lists do not intersect, so return null.
 *
 *
 * 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.
 */
public class IntersectionNode {
    public class ListNode {
        int val;
        ListNode next;
        ListNode(int x) {
            val = x;
            next = null;
        }
    }
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        ListNode dummyA = headA, dummyB = headB;
        int sizeA = 0, sizeB = 0;
        if (dummyA == null || dummyB == null) return null;
        while (dummyA != null) {
            dummyA = dummyA.next;
            sizeA++;
        }
        while (dummyB != null) {
            dummyB = dummyB.next;
            sizeB++;
        }
        while (sizeB < sizeA) {
            headA = headA.next;
            sizeB++;
        }
        while (sizeA < sizeB) {
            headB = headB.next;
            sizeA++;
        }
        while (headA != headB) {
            headA = headA.next;
            headB = headB.next;
        }
        return headA;
    }
}
