Right? If that was the case in that code, then my complexity is no longer log, because I need linear access for each time I've got to go to the list, and it's going to Lisp be much worse than that.

这里的复杂度不再是对数的了，因为每次在列表中，查找需要线性访问,可能还要糟糕,其实，有些编程语言，如。

麻省理工公开课 - 计算机科学及编程导论课程节选

You want to write the specs so that everybody does in fact know what this piece of code is doing, and you're writing it each time around.

但是你得养成良好的习惯和编程风格,你得撰写规范说明从而保证,每个人都知道这段代码是干什么的。

麻省理工公开课 - 计算机科学及编程导论课程节选

I can change everything overnight; it does happen. I put that in for a reason because sometimes some of you have not taken a physics course and you don't know how to do well in physics and slowly you catch on and by the time it's Final exam you crack the code; you know how to do well.

我能一夜之间扭转乾坤,确实有过这样的例子,我这样做是因为,有时你们中有人可能以前没上过物理课,不知道怎样学好物理,但慢慢地你赶了上来,到期末考试的时候,你忽然灵光一闪,考起试来得心应手

耶鲁公开课 - 基础物理课程节选

So really, the rule of thumb if you think back to Scratch is -- that any time you write a statement -- -- something that does something in one line of code -- you generally need the semicolon there.

实际上，根据经验，你们回忆一下在Scratch中,任何时候你们写一条语句-,处理一些操作的一行代码-,你们通常需要在那里加个分号。

哈佛公开课 - 计算机科学课程节选

I stress it in particular because I know you're all going to get into this stage; you've got a problem set due in a couple of hours, you're still writing the code, you don't want to waste time, and I'm going to use quotes on "waste time", doing those extra things to do the defensive programming, you just want to get the darn thing done.

我特别强调了这一点,因为我知道你们都会慢慢成为程序员;,你在写代码过程中造成了一系列的错误，你还在写代码，你不想浪费时间，我想把“浪费时间“用引号圈起来，浪费时间来做因为防卫性程序设计，而带来的这些额外的事情,你只想赶紧把程序写完，这不是一个好习惯，因为当你做完回头。

麻省理工公开课 - 计算机科学及编程导论课程节选

Remember last time, I said that there's different kinds of complexity in our code, and I suggested for simple branching programs, the amount of time it takes to run that program is, in essence, bounded by the number of instructions, because you only execute each instruction at most once.

但是这里有个很重要的点,记得上节课，我提过在我们的代码中，有不同种类的复杂度，而且我还说了对于简单的分支程序,运行这种程序需要的总体时间,大体上，是和指令的数目相关的,因为每个指令只会被执行最多一次。

麻省理工公开课 - 计算机科学及编程导论课程节选

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。

哈佛公开课 - 计算机科学课程节选