[Leetcode] 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):

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:

 public class Solution {
     public String getPermutation(int n, int k) {
         if(n==1&&k==1)
             return "1";
         int total=getTotal(n);
         String original=getOriginal(n);
         char[] result=new char[n];
         for(int i=0;i<n;++i){
             total/=(n-i);
             int t2=(k-1)/total;
             result[i]=original.charAt(t2);
             original=original.replace(result[i]+"", "");  //这个方法很巧妙啊，用此法就可以把用过的数字从数组里去掉了！！！
             k-=t2*total;
         }
         return new String(result);
     }

     private String getOriginal(int n) {
         // TODO Auto-generated method stub
         String result="";
         for(int i=1;i<=n;++i){
             result+=i+"";
         }
         return result;
     }

     private int getTotal(int n) {
         // TODO Auto-generated method stub
         int total=1;
         for(int i=1;i<=n;++i){
             total*=i;
         }
         return total;
     }
 }