题解 | #两个链表的第一个公共结点#

暴力解法

/*
struct ListNode {
	int val;
	struct ListNode *next;
	ListNode(int x) :
			val(x), next(NULL) {
	}
};*/
class Solution {
public:
    ListNode* FindFirstCommonNode( ListNode* pHead1, ListNode* pHead2)
	{
		if(pHead1==NULL||pHead2==NULL)
		return NULL;
		struct ListNode  * p1=pHead1;
		struct ListNode  * p2=pHead2;
		while(p1!=NULL)
		{
			p2=pHead2;//重置p2，使p2重新指向表头，开始下一次的遍历。
			while(p2!=NULL)
			{
				if(p2==p1)
				{
					return p1;
				}
				p2=p2->next;
			}
			p1=p1->next;
		}
		return NULL;
    }
};

巧妙解法

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