leetcode中Intersection of Two Linked Lists问题

##问题描述
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.

即要寻找相交节点，并且要求时间复杂度为n，空间复杂度为1

##问题分析
该问题解决方法非常巧妙
利用list中的val值，计算listB中的所有节点val的和countB
然后计算listA中val++和的值，最后再次计算listB中val的值得和newCountB
比较一下countB和newCountB，如果两者相同，说明listB没有受到listA的影响，则listA和listB无交点
否则则说明有交点，并且交点位置为lengthB - newCountB-countB

Tip：这个方法太牛逼了（不是我想到的）

##C++代码
/**

 * 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) {
        ListNode *tmp = headB;
        int countB = 0,lengthB = 0;
        while(tmp != NULL){
            countB += tmp->val;
            tmp = tmp->next;
            lengthB++;
        }

        tmp = headA;
        while(tmp != NULL){
            tmp->val++;
            tmp = tmp->next;
        }

        tmp = headB;
        int newCountB = 0;
        while(tmp != NULL){
            newCountB += tmp->val;
            tmp = tmp->next;
        }

        //默认不允许修改list的结构，所以要将其还原
        tmp = headA;
        while (tmp!=NULL)
        {
            tmp->val--;
            tmp = tmp->next;
        }

        if(newCountB == countB) return NULL;
        //应该写成if else结构的，但是实际这两种写法的效果是一样的
        tmp = headB;
        for(int i = 0;i < lengthB-(newCountB-countB);i++){
            tmp = tmp->next;
        }
        return tmp;
    }
};