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It actually lets you see how to get access to that pointer so you know what you're referring to.
怎么来取得指针,以得知在对什么进行操作,但是这是个简单的设计决定。
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On the other hand, if the value I'm looking for here- sorry, the value I'm looking for is smaller than the value I see here, I just need to look here. All right?
如果我的目标数比这个值要小呢？,我就在这边找就对了，对不对？,做完了这一步,我可以在下一步做相同的操作，假设我选中了这一分支？
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Now, suppose in fact these weren't x and y glued together, these were radius and angle glued together.
我实际也说过了,我在这里操作的是，和这两个点。
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If you give me a number and a string, I'm going to concatenate them together, it's really different operations, but nonetheless, it's what it's going to do.
如果给的输入是一个数字和一个字符串，我将会把它们连接在一起，这实在是不正确的操作，但是无论如何，它就是会去这么做。
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And those are two things that you'd like to do with every looping construct you write: you'd like to be able to assure yourself that they will always terminate, and then the second thing you'd like to do, is to assure yourself that it does give you back a reasonable answer.
就是现在我有了能去检查程序,是不是做了正确的操作的能力：,这就是你们在写每个循环程序的时候,都要注意的两个事情：,第一件事情是要确保程序可以终结，第二件你要做的事情,就是要确保程序返回了正确的答案。
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And we're going to see, well all right.
在这一块程序中做的操作。
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So we're to assume we can get to any piece of data, any instruction in constant time, and the second assumption we're going to make is that the basic primitive steps take constant time, same amount of time to compute. Again, not completely true, but it's a good model, so arithmetic operations, comparisons, things of that sort, we're all going to assume are basically in that in that particular model.
因此如果我们假设在恒定的时间内,我们可以取得任何一块数据,任何一种数据结构的话,我们要做的第二个假设就是,基本的原始操作计算花费的时间是恒定的,这个假设也不是完全正确的，但这个模型其实挺不错的，因此算法操作，比较，这一类的事情，我们在这个特定的模型中都假设是基本的,操作，花费的时间是恒定相同的。
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O Right there, order n. So I have order n operations at each level in the tree. And then how many levels deep am I? Well, that's the divide, right? So how many levels do I have?
在这儿,所以我在树的每个层次都要做O的操作,那个这棵有多少层呢？,这是一个除法，不是吗？
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