对复杂性列表进行排序（Big O）
给定一个复杂的列表：
那么你如何按照他们的 Big O 顺序排序？
我想答案就在下面？
现在的问题是 log(n!) 如何变成 n log(n)。我也不知道我是否得到了n！和 (n-1)!正确的。 c^n 有可能比 n! 更大吗？当c＞1时n？
一般来说，我如何可视化这样的大O问题...我花了很长时间才做到这一点...与迄今为止的编码相比...任何资源，视频麻省理工学院开放课件资源，带有解释的东西
Given a list of complexities:
How do you order then in their Big O order?
I think the answer is below?
Question now is how does log(n!) become n log(n). Also I don't know if I got the n! and (n-1)! right. Is it possible that c^n can be bigger than n!? When c > n?
In general how do I visualize such Big O problem ... it took me quite long to do this ... compared to coding so far ... Any resources, videos MIT Open Courseware resources, something with explaination
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您可能想看看功能如何增长。这是 Wolfram Alpha 的快速绘图：
link
一般来说，
n^n的增长速度比c^n对于任何大于某些n_0的n（因为n将超过c在某个时刻，即使 c 非常大）。对数增长比二次或指数增长慢得多，比线性增长稍快。对于
O(log(n!)) = O(nlogn)，我相信有一种叫做斯特林近似的东西。归结起来就是将O(n!) = O(n^n)视为n! = n*(n-1)*(n-2)*...*2*1，因此n^n = n*n*n*...*n是一个上限。可以证明它也是一个下界，但你不需要它。由于
log(n^n) = nlogn根据对数规则，O(log(n!) = O(log(n^n)) = O(nlogn)。You might want to see how the functions grow. Here's a quick plot from Wolfram Alpha:
link
In general,
n^ngrows much faster thanc^nfor anyngreater than somen_0(becausenwill overtakecat some point, even if c is extremely large). log grows much slower than quadratic or exponential, and slightly faster than linear.For
O(log(n!)) = O(nlogn), I believe there was something called Stirling's Approximation. It boils down to seeing thatO(n!) = O(n^n)asn! = n*(n-1)*(n-2)*...*2*1, son^n = n*n*n*...*nis an upper bound. It can be proven that is it a lower bound as well, but you don't need that.Since
log(n^n) = nlognby log rules,O(log(n!) = O(log(n^n)) = O(nlogn).