两个链表的交叉

请写一个程序，找到两个单链表最开始的交叉节点。
注意事项
如果两个链表没有交叉，返回null。
在返回结果后，两个链表仍须保持原有的结构。
可假定整个链表结构中没有循环。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    /**
     * @param headA: the first list
     * @param headB: the second list
     * @return: a ListNode
     */
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        // write your code here
        if (headA == NULL || headB == NULL) {
            return NULL;
        }
        ListNode *orgiHeadA =headA;
        ListNode *orgiHeadB =headB;
        
        ListNode *longHead;
        ListNode *shortHead;
        
        int longCount ;
        int shortCount ;
        
        int countA = 1;
        int countB = 1;
        
        while(headA = headA->next) {
          countA++;
        }
        while(headB = headB->next) countB++;
        
        if (countA > countB) {
            longCount = countA;
            shortCount = countB;
            longHead = orgiHeadA;
            shortHead = orgiHeadB;
        } else {
            longCount = countB;
            shortCount = countA;
            longHead = orgiHeadB;
            shortHead = orgiHeadA;
        }
        
        int delta = longCount - shortCount;
        while(delta) {
            longHead = longHead -> next;
            delta--;
        }
        
        while(longHead) {
            if (longHead == shortHead) {
                return longHead;
            }
            longHead = longHead->next;
            shortHead = shortHead->next;
        }
        
        return NULL;
        
    }
};

http://www.lintcode.com/zh-cn/problem/intersection-of-two-linked-lists/