Now you might say, wait a minute. Thing's ordered, if I stop part way through and I throw away half the list, doesn't that help me? And the answer is yes, but it doesn't change the complexity.

如果我在半路上停下来,然后不去遍历剩下的数组了,这会有帮助么？答案是有帮助，但这没法改变算法的复杂度，因为我们之前怎么说来着？

麻省理工公开课 - 计算机科学及编程导论课程节选

It says, well I'm going to print out first and last just so you can see it, and then I say, gee 2 if last minus first is less than 2, that is, if there's no more than two elements left in the list, then I can just check those two elements and return the answer.

然后它计算了尾点和开始点的差,如果小于2的话,也就是说数组中的元素小于等于，我对这两个元素进行比较,然后返回结果就可以了，否则的话，我们就去寻找中值点，注意它是怎么实现的,首先这个指向一个列表的开头。

麻省理工公开课 - 计算机科学及编程导论课程节选

My answer to that list is yes.

我觉得单子上的都是。

普林斯顿公开课 - 国际座谈会课程节选

It says while the index is smaller than the size - of the list, I'm not at the end of the list and I don't have an answer yet, check.

当然是如果我还没检索到目标数的话,因此我会去看看是不是,这里真的看不情。

麻省理工公开课 - 计算机科学及编程导论课程节选