leetcode中题目位置
https://leetcode.cn/problems/linked-list-cycle-ii/submissions/?envType=study-plan-v2&envId=top-100-liked
代码：
public class Solution {public ListNode detectCycle(ListNode head) {if (head == null || head.next == null) {return null;}// a + b = slow = cnt;// a + b + (b + c)x = fast = 2 * slow = 2cnt ----> bx+cx = cnt// a = b(x-1) + cx = (b + c)(x - 1) + c, 即从 相遇节点、root节点出发，经过(a+1)节点后，他们会在头结点相遇；ListNode slow = head;ListNode fast = head.next;while (slow != fast) {if (fast.next == null || fast.next.next == null) {return null;}slow = slow.next;fast = fast.next.next;}slow = slow.next;while (head != slow) {head = head.next;slow = slow.next;}return head;}
}重点：快慢指针相遇后，慢指针继续往前，同时root也开始往前（root.next = head), 他们必然会相遇，即a = (b + c)(x - 1) + c
* a+1是root到环起点的步数；
* c+1是慢指针从相遇节点 到 环起点的步数；
* b+c是环的节点数
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