So if I got these two cells they would be less they would have less potential than these.

因此这些细胞它们会,它们拥有的潜能就会比较少

耶鲁公开课 - 生物医学工程探索课程节选

If there are less than two elements there I just check one or both of those to see if I'm looking for the right thing.

列表里是否剩余超过两个元素？,是否剩余不足两个元素,这里面我就挑一两项来检查下是否运行正常。

麻省理工公开课 - 计算机科学及编程导论课程节选

So the take away is that if the user actually gives me two or three or four or whatever, this is this expression "is n less than 1" is going to evaluate to false if it's actually two or three or four or whatever.

所以先不管用户实际上给我,3，4还是其他数，这个表达式,“是否n小于1“将被评估为错误的,如果输入额是2，3，4或其它的数。

哈佛公开课 - 计算机科学课程节选

So first of all, I've got to be careful about the end test. But the second thing is, OK, if it stops whenever this is less than two, am I convinced that this will always halt?

小心最后一次比较,其次，如果少于两个元素程序停止了,我们能确信这总是应该终止的么？,答案是肯定的，因为我在做什么？

麻省理工公开课 - 计算机科学及编程导论课程节选

It says, well I'm going to print out first and last just so you can see it, and then I say, gee 2 if last minus first is less than 2, that is, if there's no more than two elements left in the list, then I can just check those two elements and return the answer.

然后它计算了尾点和开始点的差,如果小于2的话,也就是说数组中的元素小于等于，我对这两个元素进行比较,然后返回结果就可以了，否则的话，我们就去寻找中值点，注意它是怎么实现的,首先这个指向一个列表的开头。

麻省理工公开课 - 计算机科学及编程导论课程节选

So this line or these lines of code up here are arguably constant time steps to say if N is less than 2 in return, that it will always take maybe one step, maybe two steps, some number of fixed CPU cycles.

如果N小于2并返回,那么这些行所对应的代码,通常只需要执行一步，或者两步，具体数字与CPU周期有关。

哈佛公开课 - 计算机科学课程节选

So if N is not less than 2, that is I have two elements or more in which case there's definitely some sorting to be done.

因此如果N不小于2，也就是有2个或更多的元素,那当然是需要做排序的。

哈佛公开课 - 计算机科学课程节选