[LeetCode] Permutation Sequence

Problem:

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.

Solution:O(n)

public class Solution {
   public String getPermutation(int n, int k) {
      ArrayList<Integer> list = new ArrayList<>();
      for (int i = 1; i <= n; i++) {
         list.add(i);
      }
      k--;
      int mod = 1;
      for (int i = 1; i <= n; i++) {
         mod = mod * i;
      }
 
      String ret = "";
 
      for (int i = 0; i < n; i++) {
         mod = mod / (n - i);
         int curIndex = k / mod;
         k %= mod;
         ret += list.get(curIndex);
         list.remove(curIndex);
      }
      return ret;
   }
}

Key Concept:Use factorial and determine the first element and so on.