60. 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.

int factorial(int n) {    int muln = 1;    while(n) muln *= n--;    return muln;}string getPermutation(int n, int k) {    vector<int> nums;    for (int i = 1; i <= n; i++) nums.push_back(i);    string res = "";    while (!nums.empty()) {        int temp = (k - 1) / factorial(n - 1);        res += to_string(nums[temp]);        nums.erase(nums.begin() + temp);        k -= temp*factorial(n - 1);        n--;    }    return res;}