-
But he says the United States can't solve the world's problems alone and it is time for all U.N.
VOA: standard.2009.09.23
-
And we said that was log rhythmic, took log n time where n is the size of the list.
当列表的长度为n的时候，整个算法耗时log，n的时间。
-
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？
-
The U.N.climate change talks opened Monday with warnings that time was running out on reaching a deal.
VOA: standard.2009.09.28
-
So I have n operations log n times, n log n there we go, n log n. Took us a long time to get there, but it's a nice algorithm to have.
所以我log，n遍的n次操作，就得到了,虽然花了不少时间得到了这个结论。
-
Mr.Obama urged Tehran to agree to a U.N.proposal to ship its uranium abroad for processing, adding that time for diplomacy on the issue is running out.
VOA: standard.2009.11.16
-
n So the velocity is given by this product of the quantum number n Planck constant 2 pi mass of the electron time the radius of the orbit, which itself is a function of n.
速度是量子数,普朗克常数2π乘以轨道半径的值,它自身也是n的函数。
-
Thetop U.N.climate official Yvo de Boer said it is time to act.
VOA: standard.2009.12.06
-
And we talked about the equation you can use for radial nodes last time, and that's just n minus 1 minus l.
我们讲过这个用于,计算径向节点的方程，也就是ｎ减去ｌ减去１
-
Yvo de Boer,the U.N.'s top climate change official, told delegates there was still time to break deadlocks, but he said countries that could be more ambitious should act now.
VOA: standard.2009.09.28
-
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。
-
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。
-
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？
-
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。
-
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。
-
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。
-
Remember, we saw that last time looking at the binary numbers. 2 to the n is a big number.
还记得吗，我们上次看过的二进制数,从2到n是一个很大的数。
-
So in other words, every time I merge the point that I kept emphasizing verbally there and that I'm only touching each number once, means that we have to account for the amount of time it takes to merge N which is going to be just N. Now, this is again one of these cyclical answers.
换言之，之前在做合并时,我不停地强调,对每个数字我只碰了一次，这就是说,我们要记录合并所花的时间量,也就是这里的,这又是一种循环性的答案。
-
So the running time of the problem where the input is T of size N as expressed here formulaically, T of N, the running time of an algorithm, given an input of size N. You know what?
因此一个输入为N的问题的运行时间,在这儿的公式表示为，如果输入为N，那么此算法的运行时间,是多少呢？
-
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。
-
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。
-
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？
-
I think most and you are familiar with the Aufbau or the building up principle, you probably have seen it quite a bit in high school, and this is the idea that we're filling up our energy states, again, which depend on both n and l, one electron at a time starting with that lowest energy and then working our way up into higher and higher orbitals.
我认为你们大多数熟悉奥弗堡,或者构建原理，你们可能,在高中见过它，又一次,这是我们填充能级的观点，与n和l有关，一个电子每次从,最低的能级开始,然后以我们的方式上升到,更高更高的轨道。
-
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。
应用推荐