算法1

(暴力枚举) $O(n^2)$

C++ 代码

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
 /*
 a+c+b 交叉走
 b+c+a 两个是相等的
 */
class Solution {
public:
    ListNode *findFirstCommonNode(ListNode *headA, ListNode *headB) {
        auto p = headA,q = headB;
        while(q != p ){
            if( q) q = q -> next;
            else q = headA;
            if(p ) p = p -> next;
            else p = headB;
        }
        return p ;
    }
};