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: "123" "132" "213" "231" "312" "321" Given n and k , return the k th 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" class Solution { public String getPermutation(int n, int k) { int[] factorial = new int[n]; int i; StringBuilder sb = new StringBuilder(); ArrayList<Integer> num = new ArrayList<Integer>(); for (i = 0; i < n; i++) { num.add(i + 1); factorial[i] = i == 0 ? 1 : i * factorial[i - 1]; } while (n > 0) { int remaining = ((k - 1) % factorial[n - 1]) + 1; sb.appen