题解 | #判断链表中是否有环#

class Solution {
public:
	bool hasCycle(ListNode *head) {
    //链表不为空
    	if(head == nullptr || head->next == nullptr )
   	 {
        	return false;
    	}
    	ListNode* fast = head;
   	 ListNode* slow = head;
   	 while(fast)
    	  {   //快慢指针各走一步
      	    fast = fast->next;
      	  slow = slow->next;
          //只有一个值时
      	  if(!fast)
      	  {
          	  return false;
      	  }
          //快指针再走一步
        	 fast = fast->next;
             //相等即有环
        	if(fast == slow)
       	 {
       	     return true;
      	  }

   	 }
     //循环结束没有环
	    return false;
   }
};