The sword refers to the offer topic 1

1. Topic description:

Enter two linked lists and find their first common node.

class Solution {
public:
    ListNode* FindFirstCommonNode(ListNode* pHead1, ListNode* pHead2) {
        /*参考大佬*/
        if (pHead1==NULL||pHead2==NULL)
        {
            return NULL;
        }
        ListNode*p1 = pHead1;
        ListNode*p2 = pHead2;
        while (p1!=p2)
        {
            p1 = (p1 == NULL) ? pHead2 : p1->next;//第一趟走完走p2
            p2 = (p2 == NULL) ? pHead1 : p2->next;//第一场走完走p1,到第一个相同点时，走过路径的长度相等,
        }
        return p1;
    }
};

ps: Refer to the big guy's solution, the idea is very good, cleverly solve the linked list with different lengths without traversing the length of the linked list, and a similar problem is to use two pointers with different steps to find out whether it is a circular linked list ,