Second step, hm. That also looks constant, you think? Oh but wait a minute. I'm accessing s.

第二步，恩，看起来也像常量,你认为呢？等等，我在进入这个数组。

麻省理工公开课 - 计算机科学及编程导论课程节选

And as a consequence, access time in the list is constant, which is what I want.

这会占用一些额外的空间，回到问题本身，这么做的话，读取数组的时间就变成常量了。

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Don't think so, right? So, what does this suggest? Sorry?

对不起我没听清，常量？哦，常量意味着无论数组的大小？

麻省理工公开课 - 计算机科学及编程导论课程节选

With this, if I can assume that accessing the i'th element of a list is constant, then you can't see that the rest of that analysis looks just like the log analysis I did before, and each step, no matter which branch I'm taking, I'm cutting the problem down in half.

读取数组中的第i个元素,是个常量时间的操作的话，我也就能像以前那样得到,这个算法是对数级复杂度的分析,并且每一步不管我选择哪个区间，我都可以把问题的规模缩小一半。

麻省理工公开课 - 计算机科学及编程导论课程节选