OK. You can see that thing is cutting down the problem in half each time, which is good, but there's one more thing I need to deal with.

这很棒，但是我们还需要处理另外一件事情,让我们仔细来看看,我在开始做之前一直在说，应该先试验试验。

麻省理工公开课 - 计算机科学及编程导论课程节选

In fact, it took the same number of steps as it did in the other case, because each time I'm cutting it down by a half.

因为每次我都把问题的规模,缩小一半，这很棒，好，接下来让我们这么做。

麻省理工公开课 - 计算机科学及编程导论课程节选

Because I'm cutting down the problem in half at each time. You're right, but there's something we have to do to add to that, and that's the last thing I want to pick up on.

但是我们还需要强调一点,这是我最后想讲的一点,让我们来看看代码-实际上，让我们先来测试测试吧。

麻省理工公开课 - 计算机科学及编程导论课程节选

And you can see it wrapping its way down, cutting in half at each time until it gets there, but it takes a while to find. All right.

但是这也花了一点时间来搜索，好，让我们看看100万，或者1000万是不是在数组里。

麻省理工公开课 - 计算机科学及编程导论课程节选

At each stage, no matter which branch, here or here, I take, I'm cutting down the length of the list that I'm searching in half. All right?

选的是这里还是那里,我总是把列表分成两半，对吧？,所以如果我处理一个长度为n的列表？

麻省理工公开课 - 计算机科学及编程导论课程节选

Log n Log n, because at each stage I'm cutting the problem in half. So I start off with n then it's n n/2 n/4 n/8 over two n over four n over eight.

因为总共有多少层？,因为在每一层，我都是把问题分解成两半,因此以n开始，然后是。

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