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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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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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to the n, every value in the 1 bit vector we looked at last time is either 0 or 1. So it's a binary n number of n bits, 2 to the n.
从2到n，我们上次看到的,位向量的每个值不是0就是,所以它是n，比特的二进制数,从2到。
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Like what the heck have we been spending our time for-- our time on with Bubble Sort and with Selection Sort and in fact there's plenty of other N squared sorts that we're not even gonna bother looking at.
真见鬼，我们竟然在-,冒泡排序和选择排序上花时间,而事实上，还有很多我们根本都不想考虑的,复杂度为N平方的排序方法。
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So this line or these lines of code up here are arguably constant time steps to say if N is less than 2 in return, that it will always take maybe one step, maybe two steps, some number of fixed CPU cycles.
如果N小于2并返回,那么这些行所对应的代码,通常只需要执行一步，或者两步，具体数字与CPU周期有关。
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I could still do the linear case, which is order n or I could say, look, take the list, let's sort it and then search it. But in that case we said well to sort it was going to take n log n time, assuming I can do that.
我仍然可以做O的线性搜索,或者也可以以这个列表为例，我们先将其进行排序,然后再进行查找，但是在这种情况下,要花费n，log，n的时间去对其进行排序。
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On the other hand, if I want to sort it first, OK, if I want to do sort and search, I want to sort it, it's going to take n log n time to sort it, and having done that, then I can search it in log n time.
我先排序，好的，如果我想排序再搜索，我要排序，这需要花n，log，n时间排序，然后做完了，我们能花log，n时间搜索,啊，哪一种更好呢？恩，呵呵。
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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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If I'm using algorithm that I'm now calling merge sort, T the running time involved in sorting N elements, T of N, you know, is just the same as running the algorithm for the right half, plus what's this plus N come from?
如果我用归并排序算法，对N个元素其运行时间，就等于此算法一半元素的运行时间,另一半的运行时间，再加上N，这个N是什么呢？
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And then one of the things that I suggested was that if we could figure out some way to order it, and in particular, if we could order it in n log n time, and we still haven't done that, but if we could do that, then we said the complexity changed a little bit.
这就涉及到了排序,如果可以想出一种来将其进行排序,甚至可以在n，log，n的时间内完成,虽然目前我们没做这件事,但是一旦开始做这件事,那么复杂性就是发生一些变化。
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