Intersection of Two Linked Lists

class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
		if(headA==NULL ||headB==NULL)
			return NULL;
		int lenHeadA=0,lenHeadB=0;
		ListNode *p=headA,*q=headB;
		//求链表headA和headB的长度
		while(p){
			p=p->next;
			lenHeadA++;
		}
		p=headA;
		while(q){
			q=q->next;
			lenHeadB++;
		}
		q=headB;
		int distance=0;
		if(lenHeadA>lenHeadB){
			distance=lenHeadA-lenHeadB;
			for(int i=0;i<distance;i++)
				p=p->next;
		}
		if(lenHeadA<lenHeadB){
			distance=lenHeadB-lenHeadA;
			for(int i=0;i<distance;i++)
				q=q->next;
		}
		//循环结束的条件是p==q!=null(有相交点) 
		//或者是p==q==null(无相交点)
		while(p!=q){
			p=p->next;
			q=q->next;
		}
		return p;
};