intersection of two linked lists

/*
Write a program to find the node at which the intersection of two singly linked lists begins.

For example, the following two linked lists:


begin to intersect at node c1.

 

Example 1:


Input: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,0,1,8,4,5], skipA = 2, skipB = 3
Output: Reference of the node with value = 8
Input Explanation: The intersected node's value is 8 (note that this must not be 0 if the two lists intersect). From the head of A, 
it reads as [4,1,8,4,5]. From the head of B, it reads as [5,0,1,8,4,5]. There are 2 nodes before the intersected node in A; There are
3 nodes before the intersected node in B.
 

Example 2:


Input: intersectVal = 2, listA = [0,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1
Output: Reference of the node with value = 2
Input Explanation: The intersected node's value is 2 (note that this must not be 0 if the two lists intersect). From the head of A, 
it reads as [0,9,1,2,4]. From the head of B, it reads as [3,2,4]. There are 3 nodes before the intersected node in A; There are 1 node 
before the intersected node in B.
 

Example 3:


Input: intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2
Output: null
Input Explanation: From the head of A, it reads as [2,6,4]. From the head of B, it reads as [1,5]. Since the two lists do not intersect, 
intersectVal must be 0, while skipA and skipB can be arbitrary values.
Explanation: The two lists do not intersect, so return null.
*/

class Solution {
public:
    ListNode *getIntersectionNode(ListNode *A, ListNode *B) {
        if(A==NULL || B==NULL) return NULL; 
        int m=len(A);
        int n=len(B);
        if(m>n){
            int d=m-n;
            while(d--){
                A=A->next;
            }
        }
        else{
            int d=n-m;
            while(d--){
                B=B->next;
            }
        }
        ListNode *res;
        while(A && B){
            if(A==B) {
                res=A;
                break;
            }
            else{
                A=A->next;
                B=B->next;
            }
        }
return res;
    }
private:
    int len(ListNode *h){
        int c=0;
        while(h){ c++; h=h->next; }
        return c;
    }
};