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

这个题和环形链表有点异曲同工,也是在走过的距离上做文章。第一个链表走完之后走第二个链表,第二个链表走完之后走第一个链表,这样,如果有相交两个指针一定会走到一起去,而没有相交时两个指针则会同时指向 null,也正好是我们需要的结果。

/*
 * 160. Intersection of Two Linked Lists
 * https://leetcode.com/problems/intersection-of-two-linked-lists/
 * https://realneo.me/160-intersection-of-two-linked-lists/
 */
public class GetIntersectionNode {
    public static void main(String[] args) {
        ListNode head1 = new ListNode(4);
        head1.next = new ListNode(1);
        head1.next.next = new ListNode(8);
        head1.next.next.next = new ListNode(4);
        head1.next.next.next.next = new ListNode(5);
        ListNode head2 = new ListNode(5);
        head2.next = new ListNode(0);
        head2.next.next = new ListNode(1);
        head2.next.next.next = head1.next.next;

        GetIntersectionNode solution = new GetIntersectionNode();

        ListNode node = solution.getIntersectionNode(head1, head2);
        if (node != null)
            System.out.println(node.val);
    }

    private ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        ListNode nodeA = headA, nodeB = headB;

        while (nodeA != nodeB) {
            nodeA = nodeA == null ? headB : nodeA.next;
            nodeB = nodeB == null ? headA : nodeB.next;
        }

        return nodeA;
    }
}