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.

一刷
题解：
如果A,B长度相同，那么直接同步next就能找到。
如果B比A长n个点，那么当a到达尾部时，b滞后n个点。然后a从B开始，b从A开始。那么a和b剩余的长度相同。
time complexity O(n), space complexity O(1)

/**
 * 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 a = headA;
        ListNode b = headB;
        
        while(a!=b){
            a = a==null? headB : a.next;
            b = b==null? headA : b.next;
        }
        return a;
    }
}

二刷
思路同上

/**
 * 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) {
        ListNode a = headA, b = headB;
        while(a!=b){
            a = a == null? headB : a.next;
            b = b == null? headA : b.next;
        }
        return a;
    }
}