题解 | #两个链表的第一个公共结点#

/*
struct ListNode {
    int val;
    struct ListNode *next;
    ListNode(int x) :
            val(x), next(NULL) {
    }
};*/
class Solution {
  public:
    ListNode* FindFirstCommonNode( ListNode* pHead1, ListNode* pHead2) {
                // 空的判断
        if (pHead1 == nullptr || pHead2 == nullptr) {
            return nullptr;
        }
        ListNode* p1 = pHead1;
        ListNode* p2 = pHead2;
        while (p1 != p2) {
            p1 = (p1 == nullptr) ? pHead2 : p1->next;
            p2 = (p2 == nullptr) ? pHead1 : p2->next;
        }
        return p1;
    }
};