先各遍历一次看是不是交于同一node，顺便记录2个链子的长度。

然后根据长度差距再遍历找到交点。。

public class Solution {    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {        if (headA == null || headB == null) return null;        ListNode dum1 = new ListNode(0);        ListNode dum2 = new ListNode(0);        dum1.next = headA;        dum2.next = headB;        int count1 = 0;        int count2 = 0;        ListNode temp1 = dum1;        while (temp1.next != null) {            count1++;            temp1 = temp1.next;        }        ListNode temp2 = dum2;        while (temp2.next != null) {            count2++;            temp2 = temp2.next;        }        if (temp1 != temp2) return null;        if (count1 > count2) {            temp1 = dum1;            temp2 = dum2;        } else {            temp1 = dum2;            temp2 = dum1;        }                int diff = Math.abs(count1 - count2);        while (diff > 0) {            temp1 = temp1.next;            diff--;        }        while (temp1 != temp2) {            temp1 = temp1.next;            temp2 = temp2.next;        }        return temp1;            }}