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VOA: standard.2009.09.28

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。

哈佛公开课 - 计算机科学课程节选

Let's suppose n is 1000, and we're running at nanosecond speed.

假入我们一秒钟运算十亿次，我们已经看过了对数级，线性增长的。

麻省理工公开课 - 计算机科学及编程导论课程节选

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？

哈佛公开课 - 计算机科学课程节选

If I'm running at nanosecond speed, 1000 n, the size of the problem, whatever it is, is 1000, and I've got a log algorithm, it takes 10 nanoseconds to complete.

如果这个问题的规模，也就是n，是，如果这个问题是对数级的，这将会占据10纳秒的时间，你一眨眼的时间。

麻省理工公开课 - 计算机科学及编程导论课程节选

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？

哈佛公开课 - 计算机科学课程节选