Right, so s of i saying, given the list, look at the i'th element, true is it the same thing?

第2个元素，看他们是不是相等呢？,如果他们相等，好，把答案设置为？

麻省理工公开课 - 计算机科学及编程导论课程节选

Lisp The problem with that technique, while it's very general, is how long does it take me to find the i'th element of the list?

很多编程语言包括，都是用了这种技术,这个问题就是，当数组很大的时候？

麻省理工公开课 - 计算机科学及编程导论课程节选

Well I have to go to the zero-th element, and say OK, gee, to get to the next element, I've got to jump this here. And to get to the next element, I've got to jump to here, and to get to the next element, I've got to jump to here, until I get there.

再回到第0个元素，好的，然后找下一个元素，跳转到这里，然后再下一个元素，跳到了这里,然后再下一个元素,直到我找到那个元素,这样，我有方法来保存变长的元素。

麻省理工公开课 - 计算机科学及编程导论课程节选

How long does it take me to find the k'th element? Linear. Because I've got to walk my way down it. OK? So in this case, you have linear access. Oh fudge knuckle.

线性的！因为我得从头,向下走一步步走,所以这里是线性访问，哦，有问题了吧。

麻省理工公开课 - 计算机科学及编程导论课程节选

Here's the problem. How do I get to the nth- er, the k'th element in the list, in this case?

如何找到第n个元素呢-,在这里，如何找到第k个元素呢？

麻省理工公开课 - 计算机科学及编程导论课程节选

Just swapping them, right? I temporarily hold on to what's in the i'th element so I can move the i plus first one in, and then replace that with the i'th element.

交换他们，对么？,临时的保存下第i个元素,然后把第i+1个元素移进来，把i+1的位置替换为第i个元素。

麻省理工公开课 - 计算机科学及编程导论课程节选

It's looking at the i'th and the i plus first element and it's saying, gee, if the i'th element is bigger than the i'th plus first element, what's the next set of three things doing?

他在观察第i个元素,和第i+1个元素,如果第i个元素大于第i+1个元素，接下来的要做哪三件事情？

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

So I'm not going to worry whether this item is a vase or a watch or painting I'm just going to say it's the n'th element.

所以我不用担心这个物品,是花瓶还是手表或者油画,我只要说这是第n个元素就好。

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