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。

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

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

I've got to count my way down, which means that the access would be linear in the length of the list to find the i'th element of the list, and that's going to increase the complexity.

的位置并去访问，然后继续下去，也就意味着,找到数组中的第i个元素的方法,是关于数组的长度呈线性复杂度的，这回增加算法的复杂度。

麻省理工公开课 - 计算机科学及编程导论课程节选