160. Intersection of Two Linked Lists

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;
        
    }
}