为什么在计算大 O 时 O(n*n*n!) 会简化为 O((n+2)!)

Question

So I have have been reading the cracking the coding interview book and there is a problem where we're we have a function that does O(n* n* n.) work. The book then says this can be expressed by O((n+2).). It says Similarly O(n*n?)can be expressed by O((n+1).). I looked in all rules if permutations and did not find any way to logically get there . My first step was cool I have O(n^2 +n!) now what? I don't know what steps to take next .

Answer 1

You already know (I think) that n. = 1*2*3*...*n n. = 1*2*3*...*n . So n*n*n. = 1*2*3*...*n*n*n n*n*n. = 1*2*3*...*n*n*n .

As n gets really big, adding 1 or 2 to a factor has a decreasingly significant effect. I'm no specialist but what matter with O() is either the power of n or, in our case, the number in the ()! expression. Which gets us to shorten this to 1*2*3*...*n*(n+1)*(n+2)=(n+2)! .

Eventually, O(n*n*n!) can be expressed O((n+2)!) .

Answer 2

To calculatee x! you do x*(x-1)! recursively until x-1==1 so x.==(x-1)*(x-2)*...*1 is O(n.). Therefore to do x*x! we have (x-0)*(x-1)*...*1 which takes one extra call to our recursive function (but at the beginning, with large x value) ie (x+1)! iterations. Similarly, (x-0)*(x-0)*(x-1)*(x-2)*...*1==x²*x! requires (x+2)! function evaluations to compute, hence O((n+2).) efficiency.