/*
 * @lc app=leetcode id=160 lang=java
 *
 * [160] Intersection of Two Linked Lists
 *
 * https://leetcode.com/problems/intersection-of-two-linked-lists/description/
 *
 * algorithms
 * Easy (35.22%)
 * Likes:    2346
 * Dislikes: 228
 * Total Accepted:    334.3K
 * Total Submissions: 949.2K
 * Testcase Example:  '8\n[4,1,8,4,5]\n[5,0,1,8,4,5]\n2\n3'
 *
 * 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.
 *
 *
 */
/**
 * 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) {
    int lenA = 0;
    ListNode pA = headA;
    for (; pA != null; pA = pA.next) {
      lenA++;
    }

    int lenB = 0;
    ListNode pB = headB;
    for (; pB != null; pB = pB.next) {
      lenB++;
    }

    if (pA != pB) {
      return null;
    }

    pA = headA;
    pB = headB;
    int len = lenA - lenB;
    if (lenA < lenB) {
      pA = headB;
      pB = headA;
      len = lenB - lenA;
    }

    while (len > 0) {
      pA = pA.next;
      len--;
    }

    while (pA != null && pB != null) {
      if (pA == pB) {
        return pA;
      } else {
        pA = pA.next;
        pB = pB.next;
      }
    }

    return null;
  }
}

