![\"\"](\"https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/12/07/circularlinkedlist.png\")
Given a circular linked list, implement an algorithm that returns the node at the beginning of the loop.
\r\n\r\nCircular linked list: A (corrupt) linked list in which a node's next pointer points to an earlier node, so as to make a loop in the linked list.
\r\n\r\nExample 1:
\r\n\r\n\r\nInput: head = [3,2,0,-4], pos = 1\r\nOutput: tail connects to node index 1\r\n\r\n
Example 2:
\r\n\r\n\r\nInput: head = [1,2], pos = 0\r\nOutput: tail connects to node index 0\r\n\r\n
Example 3:
\r\n\r\n\r\nInput: head = [1], pos = -1\r\nOutput: no cycle\r\n\r\n
Follow Up:
\r\nCan you solve it without using additional space?
给定一个链表，如果它是有环链表，实现一个算法返回环路的开头节点。若环不存在，请返回 null。
如果链表中有某个节点，可以通过连续跟踪 next 指针再次到达，则链表中存在环。 为了表示给定链表中的环，我们使用整数 pos 来表示链表尾连接到链表中的位置（索引从 0 开始）。 如果 pos 是 -1，则在该链表中没有环。注意：pos 不作为参数进行传递，仅仅是为了标识链表的实际情况。
\n\n
示例 1：
\n\n
\n输入：head = [3,2,0,-4], pos = 1\n输出：tail connects to node index 1\n解释：链表中有一个环，其尾部连接到第二个节点。\n\n\n
示例 2：
\n\n![\"\"](\"https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/12/07/circularlinkedlist_test2.png\")
\n输入：head = [1,2], pos = 0\n输出：tail connects to node index 0\n解释：链表中有一个环，其尾部连接到第一个节点。\n\n\n
示例 3：
\n\n![\"\"](\"https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/12/07/circularlinkedlist_test3.png\")
\n输入：head = [1], pos = -1\n输出：no cycle\n解释：链表中没有环。\n\n
\n\n
进阶：
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