我在网上看了一些面试问题，关于你将如何找到，如果有一个链表循环和解决方案（Floyd的循环查找算法）是有两个指针，一个是比其他快两倍，并检查他们再见面的。

I read some interview question online about how would you find if there's a loop in a linked list, and solution (Floyd's cycle-finding algorithm) is to have two pointers, one is 2x faster than the other, and check if they meet again.

我的问题是：为什么我不能只保留一个指针不动，只需将其他指针向前每次1步

My question is: Why can't I just keep one pointer fixed, just move the other pointer forward by 1 step each time?

推荐答案

由于第一（非移动）指针可能不会说谎的内环路，所以指针永远不会满足。 （请记住，一个循环可以由该列表的一部分的。）

Because the first (non-moving) pointer might not lie within the loop, so the pointers would never meet. (Remember that a loop may consist of only part of the list.)