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。
So the running time of the problem where the input is T of size N as expressed here formulaically, T of N, the running time of an algorithm, given an input of size N. You know what?
因此一个输入为N的问题的运行时间,在这儿的公式表示为,如果输入为N,那么此算法的运行时间,是多少呢?
If I'm running at nanosecond speed, 1000 n, the size of the problem, whatever it is, is 1000, and I've got a log algorithm, it takes 10 nanoseconds to complete.
如果这个问题的规模,也就是n,是,如果这个问题是对数级的,这将会占据10纳秒的时间,你一眨眼的时间。
So what we'll do is this problem here, which is let's calculate out what the wavelength of radiation n would be emitted from a hydrogen atom if we start at the n equals 3 level and we go down to the n equals 2 level.
我们来做这个问题,让我们来计算一下,从n等于3到,等于2能级氢原子辐射的波长是多少。
Where you go from problem of size n to a problem of size n minus 1.
或者缩减了2,这都一样的,也就是把问题的规模从n变成了n-1。
And now I have a problem of size N minus 1.
现在是一个N-1大小的问题。
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