OK. Now, having said that, and I've actually got a subtlety I'm going to come back to in a second, there ought to be a better way to do this. OK?

我们已经明白了一些，一会儿要讲的微妙之处,应该有更好的方法对不对？,应该有更好的思路的？

麻省理工公开课 - 计算机科学及编程导论课程节选

And the problem is, that we want to build this abstract data type, but we'd like to basically know what kind of object is it, and what functions actually belong to it, how do we use them?

然后我对它们运行了这个方法，就是这一小段代码这个方法,这有意义么？,当然没有了，对不对？,因为你知道，当你把两个极坐标，表示的点进行相加运算的时候？

麻省理工公开课 - 计算机科学及编程导论课程节选

So I'm first going to look for something that's not in the list, I'm going to see, is minus 1 in this list, so it's going to be at the far end, and if I do that in the basic case, bam.

如果我试试第一种最基本的方法,噢，一下就完成了对不对？,因为这种方法查了下第一个元素，然后发现目标数比较下，因为目标数小于第一个元素。

麻省理工公开课 - 计算机科学及编程导论课程节选

So what would I want to do? I'd like to somehow walk down each of the digits one at a time and add them up. Ah, that's a looping mechanism, right? I need to have some way of walking through them.

去取这个数的,每个数字然后把他们加起来,啊，这是个循环机制对不对？,我得找到一个遍历它们的方法,一种简单的方法可能。

麻省理工公开课 - 计算机科学及编程导论课程节选

I was hoping for a simpler answer, which is, just run it. Which is, yeah I know, seems like a dumb thing to say, right?

我希望一个简单点儿的答案，比如说去运行它吧，这方法看起来挺笨的对不对？

麻省理工公开课 - 计算机科学及编程导论课程节选

Let's search to see though now if a million is in this list, or 10 million, whichever way I did this it must be a million, right?

不管我选哪个，数都挺大的对不对？,用嘴基本的方法，噢，花的时间有点长,好，而用二分法呢？

麻省理工公开课 - 计算机科学及编程导论课程节选

This is another way of doing exponentiation, but this one's a recursive function. All right?

但这是一种递归的方法对不对？,好，让我们看看这种方法？

麻省理工公开课 - 计算机科学及编程导论课程节选

First of all, that depends on which input I've picked. All right?

首先，这方法的结果跟我选中的,输入有关对不对？

麻省理工公开课 - 计算机科学及编程导论课程节选