解题思路：如果两个链表在某一点相交，那么那一点之后的node也都会相同，长度也相同。所以，我们先遍历获取对应每一条链表的长度，然后让长的链表先走两个链表长度之差的距离，然后再同时起步，每个节点进行对比，能不能找到相同的。
/*** 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 dummy_head1 = new ListNode();ListNode dummy_head2 = new ListNode();dummy_head1.next = headA;dummy_head2.next = headB;ListNode p = dummy_head1;ListNode q = dummy_head2;int count1 = 0;int count2 = 0;while(p.next!=null){count1++;p = p.next;} while(q.next!=null) {count2++;q = q.next;}p = dummy_head1;q = dummy_head2;while(count1>count2) {p = p.next;count1--;}while(count1<count2){q = q.next;count2--;}while(p.next!=null){p = p.next;q = q.next;if(p == q){return p;}}return null;}
}
