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) {
	}
};*/
#include <cstddef>
class Solution {
public:
	//来一个求链表长度的函数 返回值：-1-传入的链表为nullptr >0-链表的长度
	int LengthList(ListNode* pHead) const
	{
		if(pHead == nullptr) return -1;
		size_t len = 0;
		ListNode* p = pHead;
		while(p != nullptr)
		{
			++len;
			p = p->next;
		}
		return len;
	}

    ListNode* FindFirstCommonNode( ListNode* pHead1, ListNode* pHead2) {
        size_t len1 = LengthList(pHead1);
		std::size_t len2 = LengthList(pHead2);
		//这里要判断是不是传入链表有空的情况 
		if(len1== -1 || len2 == -1) return nullptr;
		
		ListNode* p = pHead1;
		ListNode* q = pHead2;
		if(len1>len2)
		{
			for(int i = 0;i != len1-len2;++i) p = p->next;
		}
		else 
		{
			for(int i = 0;i != len2-len1;++i) q = q->next;
		}
		while(q != p)
		{
			q = q->next;
			p = p->next;
		}
		return q;
    }
};