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OK? So the example I'm going to do, I'm going to search a sorted list.
来搜索目标元素，好，翻到课堂材料的第二页。
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Again. Basic premise of binary search, or at least we set it up was, imagine I have a sorted list of elements. We get, in a second, to how we're going to get them sorted, and I want to know, is a particular element in that list..
好，二分查找的基本前提，或者是我们建立二分查找的基础,我们已经有了一个排好序的元素列表，我们就需要知道如何来快速的排序,如何从列表中找到特定的元素。
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We don't seem to be doing that just yet, certainly not as badly, alright, so at this point in the story I have a sorted list of size 4.
当然现在我们不需要那样做,此时此刻,我已对整个问题中大小为4的列表排好序了。
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If we could do sort, then we saw, if we amortized the cost, that searching is a lot more efficient if we're searching a sorted list.
如果我们可以做排序，然后我们可以看到，如果我们分摊开支,在有序列表中搜索将会变得更高效。
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Well let's see. My fall back is, I could just do linear search, walk down the list one at a time, just comparing those things. OK. So that's sort of my base. But what if I wanted, you know, how do I want to get to that sorted list? All right?
我只能做线性搜索了,一次遍历一遍列表，一个一个比较，但如果我想要,那怎样得到有序的列表呢？,现在的一个问题是，我们排序之前？
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Where in the worlddid that sorted list come from?
如果我只是有一系列的元素，那怎么办呢？
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You'll also notice that this thing goes through the entire list, even if the list is sorted before it gets partway through.
你也能注意到，它始终会遍历列表,甚至列表在排序之前,就是有序的也是这样。
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This list is sorted and that is, you know, stupid to say but it's very much correct.
这个序列是有序的，虽然有点愚蠢,但却是正确的。
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This is sorted, this is sorted, how do I now make a list of size 2?
这个是有序的，这个也是有序的，我怎样才能组成一个有2个元素的列表？
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This list is sorted, this list is sorted but they could be intermingled.
这个序列是有序的，这个也是有序的,可以将它们组合起来。
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Alright, so the list is now hopefully sorted correctly and it is, in fact.
现在，整个列表已经正确地,排好序了。
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So now, I have a list that's sorted of size 2.
现在大小为2的列表已排好序了。
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And this is now consistent with my claim that I have sorted a list of size N equals 1.
这与我之前所说的是一致的,我已经将N为1的一个序列排好了序。
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I remind you, I know you're not really listening to me, but that's OK. I reminded you at the beginning of the lecture, I said, let's assume we have a sorted list, and then let's go search it.
没关系，我告诉过你在课程的开始,我们假设这是一个排好序的列表,然后才进行的搜索,那实际上有序列表从哪里来的呢？
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If I look for, say, minus 1, you might go, gee, wait a minute, if I was just doing linear search, I would've known right away that minus one wasn't in this list, because it's sorted and it's smaller than the first elements.
如果我要查找-1，你可能要怒了，呵呵，等一等，如果我用的是线性查找，我不会知道-1不在这个列表中,但是列表是排好序的,1又比第一个元素小。
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And the example I want to look at is, suppose I want to search a list that I know is sorted, to see if an element's in the list.
看看目标元素在不在数组里，也就是说我要去,检索一个有序的数组。
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This list is sorted.
这个序列也是有序的。
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So to just preface what we're going to do next time, what would happen if I wanted to do sort, and rather than in sorting the entire list at once, I broke it into pieces, and sorted the pieces, and then just figured out a very efficient way to bring those two pieces and merge them back together again?
所以为了引导下一次,我们要讲的内容,如果我想做排序,而且不是一次吧整个列表排完,会发生什么，我把它拆成小的列表,然后把各个小列表排序,接着用高效的方法再把小的列表？
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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：
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