Remember we said we design lists so that the access, no matter where it was on the list was of constant time.

记住我们说过我们创建了,一个列表它就是这么访问的,无论它在列表的哪个位置。

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Constant. Ooh, constant says, no matter what the length of the list is, I'm going to take the same amount of time.

这个算法用的时间是相同的，我不这么认为，如果我们创建一个十倍于以前大小的数组。

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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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That one's not so obvious. So let's think about this for a second. To sort a list in linear time, would say, I have to look at each element in the list at most a constant number of times.

所以让我们来思考一会,要在线性时间能排序，列表里每个元素最多被使用常数次,不一定是一次，对吧。

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。

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