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.

参考解析：http://bangbingsyb.blogspot.com/2014/11/leetcode-permutation-sequence.html

思路：先算n!。.然后k/(n-1)!＋1则为现在的第一个数字。更新k%=(n-1)!, n=n-1.

public class Solution {    public String getPermutation(int n, int k) {
            if(n<=0)        {            return "";        }        List
     
       list=new ArrayList
      
       ();        int fac=1;                for(int i=1;i<=n;i++)        {            list.add(i);            fac*=i;        }                k--; //zero based                StringBuilder sb=new StringBuilder();                while(n>0)        {            fac/=n;            sb.append(list.remove(k/fac));  //di ji zu                        k%=fac; //di ji ge                         n--;        }        return sb.toString();    }}