剑指-判断一个单向链表是否形成了环形结构

题目: 判断一个单向链表是否形成了环形结构.
思路: 定义两个指针,同时从链表头部出发,一个指针一次走一步,另一个指针一次走两步.如果走的快的指针追上了走的慢的指针,那么链表就是环形链表. 如果走的快的的走到了链表尾部都没追上第一个指针,那么链表就不是环形链表.

private static class Node {
    int value;
    Node next;
}

static boolean isRing(Node head) {
    if (head == null) {
        return false;
    }

    Node pre = head;
    Node rear = head;

    while (rear.next != null) {

        if (rear.next.next != null) {
            rear = rear.next.next;
        } else {
            return false;
        }

        if (rear.next.next == pre) {
            return true;
        }

        pre = pre.next;
    }

    return false;
}


public static void main(String[] args) {
    Node head = new Node();
    head.value = 1;
    addTail(head, 2);
    addTail(head, 3);
    addTail(head, 4);
    addTail(head, 5);
    addTail(head, 6);
    addTail(head, 7);
    addTail(head, 8);
    Node _9 = addTail(head, 9);
    addTail(head, 10);
    addTail(head, 11);
    addTail(head, 12);
    addTail(head, 13);
    addTail(head, 14);
    Node tail = addTail(head, 15);
    tail.next = _9;

    System.out.println(isRing(head));
}

private static Node addTail(Node head, int value) {
    Node node = new Node();
    node.value = value;
    if (head != null) {
        Node tmp = head;
        while (tmp.next != null) {
            tmp = tmp.next;
        }
        tmp.next = node;
    }
    return node;
}