算法6：相交链表

题目链接

解析：

完整代码：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
typedef struct ListNode ListNode;
class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        //1.两个链表的长度
       ListNode* pa = headA;
       ListNode* pb = headB;
       int sizeA = 0, sizeB =0;
       while(pa)
       {
        ++sizeA;
        pa = pa ->next;
       } 
        while(pb)
       {
        ++sizeB;
        pb = pb ->next;
       } 
       int gap = abs(sizeA - sizeB);
       ListNode* longlist = headA;
       ListNode* shortlist = headB;
       if(sizeA < sizeB)
       {
        longlist = headB;
        shortlist = headA;
       }
       while(gap--)
       {
        //让长链表先走
        longlist = longlist->next;
       }
       while(longlist)
       {
        if(longlist == shortlist)
            return longlist;   //shortlist
        longlist = longlist ->next;
        shortlist = shortlist ->next;
       }
       //跳出循环，两个链表不相等
       return NULL;
        
    }
};