And you can go ahead and tell me what you think the bond order is going to be for this molecule.
你们告诉我你觉得,这个分子的键序应该是怎样的。
So if you now reverse the story that we're telling here, what was the next line supposed to be after sorting left half?
回退到这个过程中来,在对左半部分排完序后,下一步该做什么呢?
If I look for, say, minus 1, you might go, gee, wait a minute, if I was just doing linear search, I would've known right away that minus one wasn't in this list, because it's sorted and it's smaller than the first elements.
如果我要查找-1,你可能要怒了,呵呵,等一等,如果我用的是线性查找,我不会知道-1不在这个列表中,但是列表是排好序的,1又比第一个元素小。
What would the bond order be for this bond?
这个键的键序是多少?
I have sorted with the smaller problem 1 because that smaller problem right now is of size 1 and so it's sort of obviously the case that this cup is now sorted.
对这个较小的问题我已经排好序了,因为在这个小问题中只有1个元素1,那么很明显,这个杯子已经是有序的了。
At the end of the first round, I've got the smallest element at the front. At the end of the second round, I've got the smallest two elements at the front, in fact I got all of them sorted out. And it actually runs through the loop multiple times, making sure that it's in the right form.
看看发生了什么,在第一轮结束后,我把最小的元素移到了前面,第二轮结束后,我把最小的,两个元素移到了前面,实际上,所有的元素都排好序了,实际上,这个算法运行了几次循环,确认下这是正确的形式。
Basic idea, before I even look at the code, is pretty simple. If I've got a list that is sorted, in let's call it, just in increasing order, and I haven't said what's in the list, could be numbers, could be other things, for now, we're going to just assume they're integers.
我们可以说基本的思想是很简单的,如果我有一个排好序的数组,让我们认为这个数组是递增的吧,我并没说数组里元素是什么,可能是数字,也可能是其他的东西,现在我们假设是integer类型的数字吧,最简单的方式就是这么做了:
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