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):

"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.

Solution

public class Solution {
    public String getPermutation(int n, int k) {
                k = k -1;
        List<Integer> candidates = new ArrayList<>();
        for (int i = 1; i <= n; i++) {
            candidates.add(i);
        }

        int[] factorial = new int[n + 1];
        factorial[0] = 1;
        factorial[1] = 1;

        for (int i = 2; i <= n; i++) {
            factorial[i] = factorial[i - 1] * i;
        }

        StringBuffer sb = new StringBuffer();

        for (int i = 1; i <= n; i++) {
            int currentPosition = k / factorial[n - i];
            k = k % factorial[n-i];

            sb.append(candidates.get(currentPosition));
            candidates.remove(currentPosition);
        }

        return sb.toString();
    }
}