And here I've got a choice. OK, one of the ways I could do would be the following.

在这种情况的数组下,我们需要考虑考虑读取时间了。

麻省理工公开课 - 计算机科学及编程导论课程节选

Or another way of saying it is, we're going to use as the basic steps, those operations that run in constant time, so arithmetic operations.

我们用可以在恒定时间内完成的操作,算法，比较,内存读取。

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。

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