/*** 检测单链表中是否有环*/public static boolean hasCircle(ListNode head) {if(null == head) {return false;}ListNode slow = head;ListNode fast = head;while(fast != null && fast.next != null){fast = fast.next.next;slow = slow.next;if(fast == slow){return true;//两个指针相遇
            }}return false;}

　　/***如果单链表有环，找出环的入口节点（环的连接点）。*这里先证明一个定理：碰撞点到连接点的距离=头指针到连接点的距离*假设单链表的总长度为L，头结点到环入口的距离为a，环入口到快慢指针相遇的结点距离为x，环的长度为r，*慢指针总共走了s步，则快指针走了2s步。另外，快指针要追上慢指针的话快指针至少要在环里面转了一圈多(假设转了n圈加x的距离)*得到以下关系：*s = a + x*2s = a + nr + x*=>a + x = nr*=>a = nr - x //入口点*/public static ListNode searchEntranceNode(ListNode head) {ListNode slow=head;//p表示从头结点开始每次往后走一步的指针ListNode fast=head;//q表示从头结点开始每次往后走两步的指针while(fast !=null && fast.next !=null) {slow=slow.next;fast=fast.next.next;if(slow==fast) {break;//p与q相等，单链表有环
            }}if(fast == null || fast.next == null){return null;}slow=head;while(slow != fast){slow=slow.next;fast=fast.next;}return slow;}

　　public static ListNode searchEntranceNode1(ListNode head){if(head == null || head.next == null)return null;ListNode slow = head;ListNode fast = head;while(fast != null && fast.next != null ){slow = slow.next;fast = fast.next.next;if(slow == fast){fast=slow;//相遇点slow=head;while(slow != fast){slow = slow.next;fast = fast.next;}if(slow == fast)return fast;}}return null;}

    /*** 求环的长度*/public static int countCircleSize(ListNode head) {ListNode entranceNode = searchEntranceNode1(head);//环的入口点if(entranceNode == null) {return 0;}ListNode slow = entranceNode.next;int size = 1;while(slow != entranceNode) {slow = slow.next;size++;}return size;}