题目：

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.

提示：

假设两个链表有合并的情况，那么合并部分的长度一定是一样的，在合并之前长度会有所不同，所以先求出长度，然后把长的链表向前走，让他们“在同一起跑线”，然后依次比较。

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) {
        if (!headA || !headB) return NULL;
        int lenA = 0, lenB = 0;
        ListNode *nodeA, *nodeB;
        for (nodeA = headA; nodeA; nodeA = nodeA->next, ++lenA);
        for (nodeB = headB; nodeB; nodeB = nodeB->next, ++lenB);
        if (lenB > lenA) {
            for (int i = 0; i < lenB - lenA; ++i, headB = headB->next);
        } else {
            for (int i = 0; i < lenA - lenB; ++i, headA = headA->next);
        }
        if (headA == headB) return headA;
        while (headA && headB) {
            headA = headA->next;
            headB = headB->next;
            if (headA == headB) return headA;
        }
        return NULL;
    }
};