[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;
    }
}