Permutation Sequence

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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):

"123"
"132"
"213"
"231"
"312"
"321"

Given n and k, return the k^th permutation sequence.

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

class Solution {public:void backtrace(string &ans, vector<int> &flag, vector<int> fac, string &str, int s, int k, int n){if(s == n){ans = str;return;}s++;int x;x = (k - 1) / fac[n - s];int t = 0;int i;for(i = 0; i < n; i++){if(!flag[i]){if(t >= x){flag[i] = 1;break;}t++;}}if(i == n){return;}str += ((i + 1) + '0');k = k - x * fac[n - s];backtrace(ans, flag, fac, str, s, k, n);}    string getPermutation(int n, int k) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        string ans;vector<int> fac(n, 1);fac[0] = 1;for(int i = 1; i<n; i++){fac[i] = fac[i - 1] * i;}vector<int> flag(n+1, 0);int index = 0;string str="";backtrace(ans, flag, fac, str, 0, k, n);return ans;    }};