使用两个节点。slow和fast，分别行进1步和2步。假设有相交的情况，slow和fast必定相遇；假设没有相交的情况，那么slow或fast必定有一个为null

public bool HasCycle(ListNode head) {
        // - for null node , false
        if(head == null || head.next == null){
            return false;
        }
        if(head.val != head.next.val && head.next.next == null){
            return false;
        }
        
        var slow = head; 
        var fast = head;


        while(true) {
            slow = slow.next;
            if(fast.next != null){
                fast = fast.next.next;
            }
            else{
                return false;
            }
            
            if(slow == null || slow.next == null || fast == null || fast.next == null) {
                return false;
            }
            
            if(slow.val == fast.val && slow.next.val == fast.next.val){
                return true;
            }
        }
        return false;
    }