输入两个链表，找出它们的第一个公共结点。

/*
struct ListNode {
int val;
struct ListNode *next;
ListNode(int x) :
val(x), next(NULL) {
}
};*/
class Solution {
public:
ListNode* FindFirstCommonNode( ListNode* pHead1, ListNode* pHead2) {
ListNode*p1=pHead1,*p2=pHead2;
/* while(p1!=p2)
{
p1=(p1==NULL)?pHead2:p1->next;
p2=(p2==NULL)?pHead1:p2->next;
}
return p1;*/
/*while (p1 != p2)
{
if (p1 == NULL)
{
p1 = pHead2;
}
else
{
p1 = p1->next;
}
if (p2 == NULL)
{
p2 = pHead1;
}
else
{
p2 = p2->next;
}
}
return p1;*/
//考虑了两个链表的长度
if(pHead1==nullptr || pHead2==nullptr)
return nullptr;
int len1 = 0;
int len2 = 0;
ListNode *ptr1 = pHead1;
ListNode *ptr2 = pHead2;
while(ptr1)
{
ptr1 = ptr1->next;
len1++;
}//链表一长度
while(ptr2)
{
ptr2 = ptr2->next;
len2++;
}//链表二长度
ptr1 = pHead1;
ptr2 = pHead2;
if(len1>len2)
{
int cnt = len1-len2;//长度差
while(cnt--)
{
ptr1 = ptr1->next;//让长的链表走到和短的链表的长度一样
}
}
if(len1<len2)
{
int cnt = len2-len1;
while(cnt--)
{
ptr2 = ptr2->next;
}
}
while(ptr1!=ptr2)//求相同结点
{
ptr1 = ptr1->next;
ptr2 = ptr2->next;
}
return ptr1;
}
};