AcWing 34. 链表中环的入口结点（C++）- 快慢指针

题目链接：https://www.acwing.com/problem/content/description/86/
题目如下：

class Solution {
public:
    ListNode *entryNodeOfLoop(ListNode *head) {
        ListNode* fast=head,*slow=head;
        //快慢指针，其中慢的每次只走一步，快的每次走两步
        while(fast!=NULL&&fast->next!=NULL){
            slow=slow->next;
            fast=fast->next->next;
            
            if(fast==slow){
                ListNode* index1=fast;
                ListNode* index2=head;
                
                while(index1!=index2){
                    index1=index1->next;
                    index2=index2->next;
                }
                
                return index1;
            }
        }
        
        return NULL;
    }
};