leetcode_intersection_of_two_linked_lists

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难度：Easy

解题思路：一个链表长度x+y，另一个z+y，则两个节点分别从两个链表的head出发，走到头后，走向另一个链表的头。当两者共同走x+y+z时，遇到相同节点。

代码如下：

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/**
 * 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 == NULL || headB == NULL)
            return NULL;
        ListNode *curA = headA;
        ListNode *curB = headB;
        while(true)
        {
            if(curA == curB)
                return curA;
            else
            {
                if(curA->next != NULL && curB->next != NULL)
                {
                    curA = curA->next;
                    curB = curB->next;
                }
                else if(curA->next == NULL && curB->next != NULL)
                {
                    curB = curB->next;
                    curA = headB;
                }
                else if(curB->next == NULL && curA->next != NULL)
                {
                    curB = headA;
                    curA = curA->next;                    
                }
                else
                    return NULL;
            }
        }
    }
};

运行结果：52ms，超过28.14%