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):

"123"
"132"
"213"
"231"
"312"
"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.

 1     string getPermutation(int n, int k) {
 2         vector<int> nums;
 3         vector<int> factors(1,1);
 4         for(int i = 1; i <= n; i++){
 5             nums.push_back(i);
 6             factors.push_back(factors[i - 1] * i);
 7         }
 8         
 9         string ret;
10         while(k > 0 && nums.size() > 0) {
11             int idx = (k -1) / factors[n -1];
12             ret.push_back(nums[idx] + 48);
13             nums.erase(nums.begin() + idx);
14             k = k - idx * factors[n - 1];
15             n --;
16         }
17         
18         return ret;
19     }