Permutation Sequence [LeetCode]

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.

Solutions: Caculates the index of the first number by  idx = (k -1) / (n-1)! ,  and the new k and new n, then goes to next round.

     string getPermutation(int n, int k) {
         vector<int> nums;
         vector<int> factors(,);
         for(int i = ; i <= n; i++){
             nums.push_back(i);
             factors.push_back(factors[i - ] * i);
         }

         string ret;
         while(k >  && nums.size() > ) {
             int idx = (k -) / factors[n -];
             ret.push_back(nums[idx] + );
             nums.erase(nums.begin() + idx);
             k = k - idx * factors[n - ];
             n --;
         }

         return ret;
     }