以下python函数的时间复杂度是多少？

Question

def func(n):
    if n == 1:
        return 1
    return func(n-1) + n*(n-1)

print func(5)

Getting confused. Not sure what exactly it is. Is it O(n) ?

Answer 1

Calculating the n*(n-1) is a fixed time operation. The interesting part of the function is calling func(n-1) until n is 1 . The function will make n such calls, so it's complexity is O(n) .

Answer 2

If we assume that arithmetic operations are constant time operations (and they really are when numbers are relatively small) then time complexity is O(n) :

T(n) = T(n-1) + C = T(n-2) + C + C = ... = n * C = O(n)

But the multiplication complexity in practice depends on the underlying type (and we are talking about Python where the type depends on the value). It depends on the N as N approaches infinity. Thus, strictly speaking, the complexity is equal to:

T(n) = O(n * multComplexity(n))

And this multComplexity(n) depends on a specific algorithm that is used for multiplication of huge numbers.

Answer 3

As described in other answers, the answer is close to O(n) for practical purposes. For a more precise analysis, if you don't want to make the approximation that multiplication is constant-time:

Calculating n*(n-1) takes O(log n * log n) (or O(log n)^1.58 , depending on the algorithm Python uses, which depends on the size of the integer). See here - note that we need to take the log because the complexity is relative to the number of digits.

Adding the two terms takes O(log n) , so we can ignore that.

The multiplication gets done O(n) times, so the total is O(n * log n * log n) . (It might be possible to get this bound tighter, but it's certainly larger than O(n) - see the WolframAlpha plot ).

In practice, the log terms won't really matter unless n gets very large.