leetcode: Intersection of Two Linked Lists

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.

	#include <iostream>
	
	
	using namespace std;
	
	
	 struct ListNode {
	     int val;
	     ListNode *next;
	     ListNode(int x) : val(x), next(NULL) {}
	 };
	
	class Solution {
	public:
	    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
	        if(headA== NULL || headB==NULL){
	            return NULL;
	        }		
	        int gapLength=compareLength(headA,headB);
	        //剪去长度
	        if(gapLength>0){
	            for(int i=0;i<gapLength;++i){
	               headA=headA->next;
	            }
	        }else {
	        	//负数取绝对值
	            int i = gapLength >> 31;
	            gapLength= ((gapLength ^ i) - i);
	            for(i=0;i<gapLength;++i){
	                headB=headB->next;
	            }
	        }
			//相同位置的比较
	        for(;;){
	            if(headA!=headB){
	                 if(headA->next!= NULL){
	                     headA=headA->next;
	                     headB=headB->next;
	                 }else{
	                     return NULL;
	                 }
	            } else{
	                return headA;
	            }
	        }
	
	
	    }
		//比较长度
	    int compareLength(ListNode *headA, ListNode *headB) {
	        ListNode *l=new ListNode(1);
	        l->next=headA->next;
	        for(;;){
	          if(l->next!= NULL){
	              l->next=l->next->next ;
	              ++l->val;
	          }else{
	              break;
	          }
	        }
	        l->next=headB->next;
	        --l->val;
	        for(;;){
	            if(l->next!= NULL){
	                l->next=l->next->next ;
	                --l->val;
	            }else{
	                break;
	            }
	        }
	        return l->val;
	    }
	};
	
	
	int main() {
	    Solution s;
	    ListNode *l1=new ListNode(1);
	    ListNode *l2=new ListNode(1);
	    ListNode *l3=new ListNode(1);
	    ListNode *l4=new ListNode(1);
	    ListNode *l5=new ListNode(1);
	    ListNode *l6=new ListNode(1);
	    ListNode *l7=new ListNode(1);
	    ListNode *l8=new ListNode(1);
	    ListNode *l9=new ListNode(1);
	    ListNode *l10=new ListNode(1);
	
	    l1->next=l2;
	    l2->next=l3;
	    l3->next=l4;
	    l4->next=l5;
	    l5->next=l6;
	    l6->next=l7;
	    l7->next=l8;
	    l8->next=l9;
	    l9->next=l10;
	    l10->next=NULL;
	
	
	
	    cout<<s.getIntersectionNode(l1,l6)<<endl;
	
	    return 0;
	
	}