complexity of recursive string permutation function

问题

From: Are there any better methods to do permutation of string?

what is the complexity of this function???

void permute(string elems, int mid, int end)
{
    static int count;
    if (mid == end) {
        cout << ++count << " : " << elems << endl;
        return ;
    }
    else {
    for (int i = mid; i <= end; i++) {
            swap(elems, mid, i);
            permute(elems, mid + 1, end);
            swap(elems, mid, i);
        }
    }
}

回答1:

Ignoring the print, the recurrence relation satisfied is

T(n) = n*T(n-1) + O(n)

If G(n) = T(n)/n! we get

G(n) = G(n-1) + O(1/(n-1)!)

which gives G(n) = Theta(1).

Thus T(n) = Theta(n!).

Assuming that the print happens exactly n! times, we get the time complexity as

Theta(n * n!)

回答2:

Without looking too deeply at your code, I think I can say with reasonable confidence that its complexity is O(n!). This is because any efficient procedure to enumerate all permutations of n distinct elements will have to iterate over each permutation. There are n! permutations, so the algorithm has to be at least O(n!).

Edit:

This is actually O(n*n!). Thanks to @templatetypedef for pointing this out.

回答3:

long long O(int n)
{
    if (n == 0)
        return 1;
    else 
       return 2 + n * O(n-1);
}

int main()
{
    //do something
    O(end - mid);
}

This will calculate complexity of the algorithm.

Actualy O(N) is N!!! = 1 * 3 * 6 * ... * 3N

来源：https://stackoverflow.com/questions/5363619/complexity-of-recursive-string-permutation-function