快慢指针 （fast and slow pointer）

public class CycleLinkedListTest {

    public static class ListNode {
        String val;
        ListNode next;

        public ListNode(String val) {
            this.val = val;
        }
    }

    public static  boolean hasCycle(ListNode head) {

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

        ListNode slow = head;
        ListNode fast = head;

        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
            if (slow == fast) {
                return true;
            }
        }

        return false;
    }

    public static ListNode detectCycleEntry(ListNode head) {
        ListNode slow = head;
        ListNode fast = head;

        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;

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

    public static void main(String[] args) {
        // 构建链表 A→B→C→D→B
        ListNode A = new ListNode("A");
        ListNode B = new ListNode("B");
        ListNode C = new ListNode("C");
        ListNode D = new ListNode("D");

        A.next = B;
        B.next = C;
        C.next = D;
        // 形成环
        D.next = B;

        // 判断是否有环
        boolean hasCycle = hasCycle(A);
        System.out.println("是否有环：" + hasCycle); // true

        // 找环入口
        ListNode entry = detectCycleEntry(A);
        if (entry != null) {
            System.out.println("环的入口节点是：" + entry.val); // B
        }
    }
}

2(a + b) = a + b + n(b+c)
化简：a = n(b+c) - b
      a = (n-1)(b+c) + c