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.

Credits:
Special thanks to @stellari for adding this problem and creating all test cases.

/**

 * Definition for singly-linked list.

 * 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 l1 = findLen(headA);

        int l2 = findLen(headB);

        if(l1>l2) {

            return findNode(headB, headA, l2, l1);

        }else {

            return findNode(headA, headB, l1, l2);

        }

        

    }

    

    private ListNode findNode(ListNode headA, ListNode headB, int l1, int l2) {

        int gap = l2 - l1;

        ListNode p1 = headA;

        ListNode p2 = headB;

        while(gap>0) {

            p2 = p2.next;

            gap--;

        }

        while(p1!=null) {

            if(p1==p2) return p1;

            else {

                p1 = p1.next;

                p2 = p2.next;

            }

        }

        return null;

    }

    

    private int findLen(ListNode node) {

        if(node==null) return 0;

        ListNode p = node;

        int l1 = 0;

        while(p!=null) {

            p = p.next;

            l1 ++;

        }

        return l1;

    }

}

==========

/**
 * Definition for singly-linked list.
 * 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;
        ListNode node1 = headA;
        ListNode node2 = headB;
        while (node1 != node2) {
            node1 = node1.next;
            node2 = node2.next;
            if (node1 == node2) return node1; // in case node1 == node2 == null
            if (node1 == null) node1 = headB;
            if (node2 == null) node2 = headA;
        }
        return node1;
    }
}