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 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.

• Given k will be between 1 and n! inclusive.

Example 1:

Input: n = 3, k = 3
Output: "213"

Example 2:

Input: n = 4, k = 9
Output: "2314"

public class Solution {
    public String getPermutation(int n, int k) {
        List<Integer> numbers = new ArrayList<>();
        int[] factorial = new int[] { 1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880 };
        StringBuilder str = new StringBuilder();
        // create a list of numbers to get indices
        for (int i = 1; i <= n; i++) {
            numbers.add(i);
        }
        k--;
        for (int i = 1; i <= n; i++) {
            int index = k / factorial[n - i];
            str.append((char) (numbers.get(index) + '0'));
            numbers.remove(index);
            k -= index * factorial[n - i];
        }
        return str.toString();
    }
}