输入两个链表，找出它们的第一个公共结点

解题思路：
要找到相交结点，首先计算出两个链表的长度，先让长的链表走abs（lenA-lenB）步，然后再同步进行遍历，如果两个结点相等则返回true否则为false。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
 typedef struct ListNode Node;
struct ListNode *getIntersectionNode(struct ListNode *headA, struct ListNode *headB) {
    Node* cur1 = headA,*cur2 = headB;
    int lenA = 0, lenB = 0;
    while(cur1)
    {
        ++lenA;
        cur1 = cur1->next;
    }
    while(cur2)
    {
        ++lenB;
        cur2 = cur2->next;
    }
    int gap = abs(lenA - lenB);
    Node* longlist = headA,*shortlist = headB;
    if(lenA < lenB)
    {
        longlist = headB;
        shortlist = headA;
    }

    while(gap--)
    {
        longlist = longlist->next;
    }

    while(shortlist && longlist)
    {
        if(shortlist == longlist)
        {
            return longlist;
        }
        else
        {
            shortlist = shortlist->next;
            longlist = longlist->next;
        }
    }
    return NULL;
}