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/*
* @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 (57.40%)
* Likes: 14737
* Dislikes: 1327
* Total Accepted: 1.5M
* Total Submissions: 2.7M
* Testcase Example: '8\n[4,1,8,4,5]\n[5,6,1,8,4,5]\n2\n3'
*
* Given the heads of two singly linked-lists headA and headB, return the node
* at which the two lists intersect. If the two linked lists have no
* intersection at all, return null.
*
* For example, the following two linked lists begin to intersect at node c1:
*
* The test cases are generated such that there are no cycles anywhere in the
* entire linked structure.
*
* Note that the linked lists must retain their original structure after the
* function returns.
*
* Custom Judge:
*
* The inputs to the judge are given as follows (your program is not given
* these inputs):
*
*
* intersectVal - The value of the node where the intersection occurs. This is
* 0 if there is no intersected node.
* listA - The first linked list.
* listB - The second linked list.
* skipA - The number of nodes to skip ahead in listA (starting from the head)
* to get to the intersected node.
* skipB - The number of nodes to skip ahead in listB (starting from the head)
* to get to the intersected node.
*
*
* The judge will then create the linked structure based on these inputs and
* pass the two heads, headA and headB to your program. If you correctly return
* the intersected node, then your solution will be accepted.
*
*
* Example 1:
*
*
* Input: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,6,1,8,4,5], skipA =
* 2, skipB = 3
* Output: Intersected at '8'
* 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
* [5,6,1,8,4,5]. There are 2 nodes before the intersected node in A; There are
* 3 nodes before the intersected node in B.
* - Note that the intersected node's value is not 1 because the nodes with
* value 1 in A and B (2^nd node in A and 3^rd node in B) are different node
* references. In other words, they point to two different locations in memory,
* while the nodes with value 8 in A and B (3^rd node in A and 4^th node in B)
* point to the same location in memory.
*
*
* Example 2:
*
*
* Input: intersectVal = 2, listA = [1,9,1,2,4], listB = [3,2,4], skipA = 3,
* skipB = 1
* Output: Intersected at '2'
* 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 [1,9,1,2,4]. 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: No intersection
* 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.
*
*
*
* Constraints:
*
*
* The number of nodes of listA is in the m.
* The number of nodes of listB is in the n.
* 1 <= m, n <= 3 * 10^4
* 1 <= Node.val <= 10^5
* 0 <= skipA < m
* 0 <= skipB < n
* intersectVal is 0 if listA and listB do not intersect.
* intersectVal == listA[skipA] == listB[skipB] if listA and listB
* intersect.
*
*
*
* Follow up: Could you write a solution that runs in O(m + n) time and use
* only O(1) memory?
*/
// @lc code=start
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*
* 思路1：两个链表若有交点，从链表末尾向前看，必有一段公共节点，因此，
* 先求出2个链表长度差gap, 长链表指针先走gap步长，直到长短两个链表位置对齐，
* 同时移动长短链表的指针，直到两个指针相等，return，
* 否则return null
*
*/
// public class Solution {
// public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
// int lenA =0;
// int lenB =0;
// ListNode nodeA = headA;
// ListNode nodeB = headB;
// while (nodeA != null) {
// nodeA = nodeA.next;
// lenA +=1;
// }
// while (nodeB != null) {
// nodeB = nodeB.next;
// lenB +=1;
// }
// nodeA = headA;
// nodeB = headB;
// if(lenB > lenA){
// int tmplen = lenA;
// lenA = lenB;
// lenB = tmplen;
// ListNode nodetmp = nodeA;
// nodeA = nodeB;
// nodeB = nodetmp;
// }
// int gap = lenA -lenB;
// while (gap-- > 0) {
// nodeA = nodeA.next;
// }
// while (nodeA != null) {
// if(nodeA == nodeB){
// return nodeA;
// }
// nodeA = nodeA.next;
// nodeB = nodeB.next;
// }
// return null;
// }
// }
/* 思路2：合并链表，同步移动指针寻找公共节点
本质上和思路1是一样的，就是指向短链表的快指针先走，
直到走完（两个链表长度之差）gap步长后， 这时快慢指针同步移动，再寻找公共节点
*
*
*/
public class Solution {
public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
ListNode pA = headA;
ListNode pB = headB;
while(pA != pB){
if(pA == null){
//指针pA走一步，直到走到a链表末尾，则转到b链表
pA = headB;
} else {
pA = pA.next;
}
//指针pB走一步，直到走到B链表末尾，则转到A链表
if(pB == null){
pB = headA;
} else{
pB = pB.next;
}
}
return pA;
}
}
// @lc code=end
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