问题
I am trying to solve this recurrence
T(n) = 3 T(n/2) + n lg n ..
I have come to the solution that it belongs to masters theorem case 2 since n lg n is O(n^2)
but after referring to the solution manual i noticed this solution that they have
The soluttion says that n lg n = O ( n ^(lg 3 - e)) for e between 0 and 0.58
so this means n lg n is O(n) .. is this right? Am i missing something here?
Isn't nlgn O(n^2) ?
回答1:
This will explain things better
回答2:
n*log(n) is not O(n^2). It's known as quasi-linear and it grows much slower than O(n^2). In fact n*log(n) is less than polynomial.
In other words:
O(n*log(n)) < O(n^k)
where k > 1
In your example:
3*T(2n) -> O(n^1.585)
Since O(n^1.585) is polynomial and dominates O(n*log(n)), the latter term drops off so the final complexity is just O(n^1.585).
回答3:
nlg3 is not O(n). It outgrows O(n)... In fact, any exponent on n that is larger than 1 results in an asymptotically longer time than O(n). Since lg(3) is about 1.58, as long as you subtract less than .58 from the exponent it is asymptotically greater than O(n).
来源：https://stackoverflow.com/questions/7830727/n-log-n-is-on