Permutation Sequence

Question:

The set [1,2,3,…,n] contains a total of n! unique permutations.

By listing and labeling all of the permutations in order,
We get the following sequence (ie, for n = 3):

1. "123"
2. "132"
3. "213"
4. "231"
5. "312"
6. "321"

Given n and k, return the kth permutation sequence.

Note: Given n will be between 1 and 9 inclusive.

Solution:

string itoax(int num)
{
    stringstream result;

    result<<num;
    return result.str();
}
int fac(int n)
{
    if(1==n || 0==n)
        return 1;
    else
        return n*fac(n-1);
}
string get_each_char(vector<int> &a,int &k)
{
    int n=a.size();
    if(1==n) 
    {
        string b=itoax(a[0]);
        a.erase(a.begin(),a.end());
        return b;
    }
    int nn=fac(n-1);
    int i=ceil(double(k)/nn);
    string b=itoax(a[i-1]);
    k=k-nn*(ceil(double(k)/nn)-1);
    a.erase(a.begin()+i-1,a.begin()+i);    
    return b;
}
class Solution {
public:
    string getPermutation(int n, int k) {
        string result;
    vector<int> a;
    for(int i=1;i<=n;i++)
        a.push_back(i);
    while(!a.empty())
    {
        string c=get_each_char(a,k);
        result +=c;        
    }
    return result;
        
    }
};