And if I'm walking down the list, this is probably order of the length of the list s because I'm looking at each element once.

这可能大概就是数组的长度，因为我会遍历数组中的每个元素一次,现在你可能会想，等等，数组已经排好序了。

麻省理工公开课 - 计算机科学及编程导论课程节选

In order, prior to lighting your cigarette you can ask your friends, "Do you mind if I smoke?"

在点燃你的烟之前，出于礼貌，你可以询问你的朋友，“Do you mind if I smoke?”

Do you mind 课堂 - SpeakingMax英语口语达人

if I could type, in order to just to get into that, in order to get into the habit.

如果我可以输入，为了不在这里陷入，为了养成这个习惯。

麻省理工公开课 - 计算机科学及编程导论课程节选

You got an extra hint here in terms of the order, so even if you had just forgotten what I said, sometimes it's not a terrible idea just to put it in the order it's written, that can give you a lot of clues as well.

其实大家还可以发现一个额外的提示，因此即使你忘了我刚才说的，有时候直接按照这个顺序,排列也不失为一个办法，这也可以给你很多提示。

麻省理工公开课 - 化学原理课程节选

I want you to talk about them. But I don't expect you to become an English major in order to do that if you're not already one.

我不指望你因为要读这些而主修英语,如果你们现在主修的并不是英语的话。

耶鲁公开课 - 1945年后的美国小说课程节选

So if we differentiate this object, I'm gonna find a first order condition in a second.

想要求它的导数,先让我想想一阶条件

耶鲁公开课 - 博弈论课程节选

Really there's no work to be done if I am handed all in sorted order so, you know, There's no work to be done if I'm handed all of the arrays in sorted order so, you know, if I demand that you give me this assumption that the cups are already sorted and then I'll sort them for you, I mean, this is kind of a cyclical argument.

如果杯子是有序排列的,那就没必要再对它进行排序了，同样如果给出的序列本身就是有序的,那也不必再做什么，如果给出这样的假设：,杯子已经有序，但仍需要对其进行排序，这像是个循环的论点。

哈佛公开课 - 计算机科学课程节选

And so I went through the math on this and said suppose I wanted to be really sloppy and I wanted to say if the delta X, the uncertainty in position is on the order of one angstrom.

我算完它,并说假如我想很草率,并且我想说如果x的增量，即位置的不确定度,相当于一埃。

麻省理工公开课 - 固态化学导论课程节选

So what I want to tell you is we also always get the same bond order if we instead only deal with the valence electrons.

我想要说的是我们如果,只考虑价电子也可以得到相同的键序。

麻省理工公开课 - 化学原理课程节选

Basic idea, before I even look at the code, is pretty simple. If I've got a list that is sorted, in let's call it, just in increasing order, and I haven't said what's in the list, could be numbers, could be other things, for now, we're going to just assume they're integers.

我们可以说基本的思想是很简单的,如果我有一个排好序的数组，让我们认为这个数组是递增的吧，我并没说数组里元素是什么，可能是数字，也可能是其他的东西，现在我们假设是integer类型的数字吧，最简单的方式就是这么做了：

麻省理工公开课 - 计算机科学及编程导论课程节选

The order complexity here, if I actually write it would be-- sorry, order n times m, and if m was equal to n, that would be order n squared, and this is quadratic.

如果m等于n的话,也就是n的平方，这是一个平方复杂度的问题，这是和前面不同类型的，好，我在做什么呢？

麻省理工公开课 - 计算机科学及编程导论课程节选

And then one of the things that I suggested was that if we could figure out some way to order it, and in particular, if we could order it in n log n time, and we still haven't done that, but if we could do that, then we said the complexity changed a little bit.

这就涉及到了排序,如果可以想出一种来将其进行排序,甚至可以在n，log，n的时间内完成,虽然目前我们没做这件事,但是一旦开始做这件事,那么复杂性就是发生一些变化。

麻省理工公开课 - 计算机科学及编程导论课程节选