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Therefore, for simple branching programs, the length of time, the complexity the code, is what we would call constant.
因此，对于简单的分支程序，运行的时间长度，算法的复杂度，也就是我们说的常数。
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But that's also nice, it lets you see how the recursive thing is simply unwrapping but the complexity in terms of the amount of time it takes is going to be the same.
它让我们看到了，在复杂度依照时间总数来看,没有变化的情况下,递归是怎么一步步的展开的,我欠你一个糖，谢谢。
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It at least does corroborate the claim that merge sort N*log N as we argue intuitively is in fact, N log N in running time.
但这至少证实了归并排序,的时间复杂度为。
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All right? It's now something that I can search in constant time. And that's what's going to allow me to keep this thing as being log.
在固定的时间内搜索,这样就可以让时间复杂度保持在对数级,好的，考虑过了这些。
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The message I'm trying to get to here, because I'm running you right up against time, is I have to be careful about what's a primitive step.
我想说的事情是，因为我正在跟大家讲算法时间复杂度，我们需要注意一个基本步骤的定义,如果我可以假设。
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I ask you for the running time of this algorithm and you give me the running time in terms of the running time, right.
我需要得到此算法的时间复杂度,那就明确地给出其,运行时间。
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That is another way of saying that looking up this thing here is constant.
都是常数级的复杂度,另一种说法就是在这里查找时间是固定的。
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And so, I get some power. I get the ability to store arbitrary things, but what just happened to my complexity?
但是复杂度变了,找到第k个元素,要花多少时间呢？
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I just go down the list selecting the smallest person at a time and then I repeat, repeat, repeat but when we actually did out the math or kind of reason through it, the running time, the asymptotic running time of bub-- of Selection Sort was also what?
只需要遍历列表，每次找出最小的元素,然后重复上述步骤，但从数学角度看,选择排序的时间复杂度,又是多少呢？
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You can actually say big O of 1, big O of 1 being constant time, the same number of steps.
其复杂度为O，表示时间是一个常量，所用的步数是相同的。
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N log N is not nearly as good as log N. As a sanity check, what algorithm have we seen that runs in log N time?
而N，log，N和log，N并不一样，我们之前探讨过的哪个算法其时间复杂度是log，N呢？
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But that merging process only takes N steps, N*log N so that's N times log of N. Now, it's a little tricky to reason through this perhaps the first time, let's just take a very simple example and see if we can do a little sanity check here.
但这个合并过程只需要N步，所以时间复杂度是,第一次对此进行推论可能会有点儿棘手，我们举一个简单的例子,看看我们能否做一些完整性的检查。
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Now, it's also the case that this is fundamentally what class this algorithm falls into, it is going to take exponential amount of time.
哪个种类的一个实例,这个问题的时间复杂度是指数级的，也就是当n上升的时候。
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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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