How many binary trees can find that haven Node and the height of these tress are n-3?
有多少二叉树可以发现有n个节点和这些枝条的n - 3的高度吗?
So, what we have left in our equation is only one part that n we haven't explained yet, and that is that n value.
现在方程里只剩下一个常数,我们还没有解释,这就是。
And then one of the things that I suggested was that if we could figure out some way to order it, and in particular, if we could order it in n log n time, and we still haven't done that, but if we could do that, then we said the complexity changed a little bit.
这就涉及到了排序,如果可以想出一种来将其进行排序,甚至可以在n,log,n的时间内完成,虽然目前我们没做这件事,但是一旦开始做这件事,那么复杂性就是发生一些变化。
And that's just a way of reminding you that we want to think carefully, but what are the things we're trying to measure when we talk about complexity here? It's both the size of the thing and how often are we going to use it? And there are some trade offs, but I still haven't said how I'm going to get an n log n sorting algorithm, and that's what I want to do today.
这只是在提醒你们我们要仔细的思考问题,但是当我们在讨论复杂性的时候,我们到底要衡量哪些东西?,是列表的大小和对其进行查找的频率吗?,这里面临一些取舍,但是我还没有说明,怎样得到一个n,log,n复杂度的排序算法,并且这是我今天想要讲的内容。
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