https://leetcode-cn.com/problems/linked-list-cycle/solution/

快慢指针

public boolean hasCycle(ListNode head) {    
        if (head == null) {
            return false;
        }
       ListNode fast = head.next;
        ListNode slow = head;
        while (fast != slow) {
            if (fast == null || fast.next == null) {
                return false;
            }
            fast = fast.next.next;
            slow = slow.next;
        }
        return true;   
    }

若还要你返回第一个入环的节点的话，那就在它们相遇时，slow停在原地， fast回到原点，变成两个都一次走一步，那么它俩一定会在第一个入环节点相遇

 public ListNode getFirstLoopNode(ListNode head) {
        if (head == null || head.next == null || head.next.next == null) {
            return null;
        }
        ListNode fast = head.next;
        ListNode slow = head;
        while (fast != slow) {
            if (fast == null || fast.next == null) {
                return null;
            }
            fast = fast.next.next;
            slow = slow.next;
        }
        fast = head;
        while (slow != fast) {
            slow = slow.next;
            fast = fast.next;
        }
        return slow;
    }