【LeetCode】详解环形链表141. Linked List Cycle Given a linked list, determine if it has a cycle in it. To

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前言

今天这道题目的第一种解法很奇葩，用计时器竟然可以AC，并且可以自己调整时间多少，跟我一起来看看吧。

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正文

原题：

链接：环形链表

Given a linked list, determine if it has a cycle in it.
To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in the linked list.
Example 1:
Input: head = [3,2,0,-4], pos = 1
Output: true
Explanation: There is a cycle in the linked list, where tail connects to the second node.

Example 2:
Input: head = [1,2], pos = 0
Output: true
Explanation: There is a cycle in the linked list, where tail connects to the first node.

Example 3:
Input: head = [1], pos = -1
Output: false
Explanation: There is no cycle in the linked list.

Follow up:
Can you solve it using O(1) (i.e. constant) memory?

题目大意
给定一个链表，判断是不是环形链表，没错就是这么简单。

思路1：

一般我们遍历链表（非环形链表）的时候，直接上来一个while (head != null)，然后循环体里面head = head.next，类似下面的语句

while (head != null) {
   
	head = head.next;
}

但是环形链表就不行了，可能会遇到无限循环导致超时的问题，这时就产生了第一种思路。
既然环形链表会超时，那我们把时间限制在某个范围之间，比如0.1s后还在运行的话，那可能就是存在环形链表了；可以使用Java自带的System.currentTimeMillis()方法获取当前时间，当然别的语言也同样提供了类似的方法，通过以下代码就可以计算出代码运行时间：

// 过去的时间
long oldTime = System.currentTimeMillis();
while (true) {
   
	// 这里是循环体代码
}
// 当前时间
long curTime = Sys