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 {
public:
string getPermutation(int n, int k) {
    // Note: The Solution object is instantiated only once and is reused by each test case.
        
	if(n == 1)
		return "1";
	char *A = new char[n];
	int mul = 1;
	for(int i = 0 ; i < n ; i ++){
		A[i] = (char)(i + '1');
		mul *= (i +1);
	}
	k = k -1;
	mul /= n;
	for(int i = 0 ; i < n -1; i ++){
		int temp = k / mul;
		k %= mul;
		mul /= (n-1 -i);
		char c= A[i + temp];
		for(int j = temp ; j >= 0; j --){
			A[i + j ] = A[i +j-1];
		}
		A[i]= c;
	}
	return string(A,A+n);
}
};