题解

快慢指针找到相遇的点

从head和相遇点位置相同速度前进，将在环入口处相遇

/**
* Definition for singly-linked list.
* class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) {
*         val = x;
*         next = null;
*     }
* }
*/
public class Solution {
 public ListNode detectCycle(ListNode head) {
     if (head == null || head.next == null || head.next.next == null) {
         return null;
     }

     ListNode slow = head;
     ListNode fast = head;
     do {
         slow = slow.next;
         fast = fast.next.next;
     }
     while (fast != null && fast.next != null && slow != fast);

     if (fast == null || fast.next == null) {
         return null;
     }

     ListNode first = head;
     ListNode second = slow;
     while (first != second) {
         first = first.next;
         second = second.next;
     }
     return first;
 }
}