Permutation Sequence

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.

class Solution {
private:
    inline  int perm( int n)
    {
         int a= 1;
         for( int i= 1;i<=n;i++)
            a*=i;
         return a;
    }
public:
     string getPermutation( int n,  int k) 
    {
         int* num= new  int[n];
         for( int i= 0;i<n;i++)
            num[i]=i+ 1;
         int dep= 1;
         string s= "";
         while(dep<=n)
        {
             int small= 1;
             int part=perm(n-dep);
             while(k>part)
            {
                k=k-part;
                small++;
            }
            s=s+ char(num[small- 1]+ ' 0 ');
             for( int i=small- 1;i<n-dep;i++) num[i]=num[i+ 1];
            dep++;
        }
         return s;
    }
};