leetcode #160 in cpp

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

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.

Code:

/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
if(!headA || !headB) return NULL;
ListNode *tempA = headA;
ListNode *tempB = headB;
int countA = 0;
int countB = 0;
//count length of A list
while(tempA->next){
countA++;
tempA = tempA->next;
}
//count length of B list
while(tempB->next){
countB++;
tempB = tempB->next;
}
//if they have intersection, tempA and tempB would reach at the same end.
//otherwise they have no intersection.
if(tempB != tempA) return NULL;
tempA = headA;
tempB = headB;
//extra_len is how much longer is a list than the other list.
int extra_len;
//if A list is longer by x
if(countA >= countB){
extra_len = countA-countB;
//we move the pointer in list A to xth node from head
while(extra_len > 0){
tempA = tempA->next;
extra_len --;
}
}else{//if B list is longer by x
extra_len = countB-countA;
//we move the pointer in list B to xth node from head
while(extra_len > 0){
tempB = tempB->next;
extra_len --;
}
}
//move both pointers at the same time. They would meet each other at the intersection node.
while(tempA != tempB){
tempA = tempA->next;
tempB = tempB->next;
}
return tempA;
}
};