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.

 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     }