"Leetcode" permutation Sequence

Title Link: https://leetcode.com/problems/permutation-sequence/

Topic:

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):

2. "123"
4. "132"
6. "213"
8. "231"
10. "312"
12. "321"

Given N and K, return the kth permutation sequence.

Note: Given n would be between 1 and 9 inclusive.

Ideas:

Calculate from left to right, the highest bit is calculated first. should be for k/(n-1)! , the next one should be (k% (n-1)!) /(N-2)!, and so on.

Algorithm:

Public String getpermutation (int n, int k) {          int nums[] = new Int[n];          int pcount=1;          for (int i=0;i<nums.length;i++) {              nums[i] = i+1;              Pcount *= (i+1);//n!          }          k--;//subscript starting from 0          String res = "";          for (int i=0;i<n;i++) {              pcount/=n-i;//(n-i-1)!                            int r = K/pcount;              res+=nums[r];//r small element for              (int j=r;j<n-1-i;j++) {//Get sub R small element so the already used element is overwritten                    nums[j] = nums[j+1];              }              K = K%pcount;          }          return res;      }  

"Leetcode" permutation Sequence