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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。
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And when I'm done with the whole thing, before I end the increment of the variable, you know, when I'm done, I'm just not returning anything out.
在循环外面初始化它，然后做一个终结测试，然后在循环内部有一个指令集，在这个例子中，就是对余数做检查，然后显示信息，当我完成了整个循环。
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This coda suggests something like a victory over all of those worries of unpreparedness as the worried rhetoric of "not yet there" gives way to a much more confident language of present completion, the "now."
这个终结暗示着,对所有的尚未准备好的忧虑的完全克服,一种对现在已经完成的事业的,自信和满足“
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When you are done you have three electron pairs in bonding orbitals.
当你完成的时候,成键轨道上共有三对电子。
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Now that seems to me a crucial understanding everything that he does That ending with the myth of the past is what he examines for the sake of articulating that.
现在对我来说很关键的是,去理解他做的一切,完成他对过去之谜探求的事业,也就是他为了阐明那一点所做的审视。
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We began here so at this point in the story the next thing that's going to happen is this line called Print F that says swap exclamation point, right?
我们从这里开始，在这里,下一步将要发生了的事情是,这行叫做printf的代码,打印：“交换完成！“对不？
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You're all going to be great chefs by the time you're done here. All right?
你们完成这门课后,都会成为大厨对不对？
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So when I'm done, I'm going to give the value back out. All right?
当它完成计算后,需要返回个值对不对？
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And one is the third element. And just return, it's done. It is a palindrome. That make sense.
完成，数组时回文的，这就对了，因为数组1a1从前面和从后面读都是一样的。
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Once I have it sorted I can search it in log n time, but that's still isn't as good as just doing n. And this led to this idea of amortization, which is I need to not only factor in the cost, but how am I going to use it?
一旦对其完成排序,就可以在log，n的时间内对其完成搜索,但是这样做仍然不如n的复杂度,这样做引出了耗时分摊的想法,这时不仅需要考虑耗时的因素？
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