Boy, there's a dumb question, because I've been telling you n log n for the last two lectures the complexity is n log n, but let's see if it really is.

孩子们，这是一个愚蠢的问题，因为前两节课的时候我就已经告诉你们了,复杂度是,但是让我们来看一下是不是真的是这样。

麻省理工公开课 - 计算机科学及编程导论课程节选

In the linear case, meaning in the unsorted case what's the complexity of this? k times n, right? Order n to do the search, and I've got to do it k times, so this would be k times n.

复杂度是多少？k的n次方，对吧？,在序列n中做搜索，要做k次，所以是k的n次方次,如果先排序后搜索。

麻省理工公开课 - 计算机科学及编程导论课程节选

The order complexity here, if I actually write it would be-- sorry, order n times m, and if m was equal to n, that would be order n squared, and this is quadratic.

如果m等于n的话,也就是n的平方，这是一个平方复杂度的问题，这是和前面不同类型的，好，我在做什么呢？

麻省理工公开课 - 计算机科学及编程导论课程节选