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 th, return the kth permutation sequence.

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

我剛開始嘗試著把，所有的全排列都列出來，然后，找到對應位置的內容，但是這樣效率太低，直接超時了，上網差了一下，和數學有關系。。。這題打算先放一放，放一下別人的代碼吧。

public String getPermutation(int n, int k) {
            char[] nums = new char[]{'1','2','3','4','5','6','7','8','9'};
            String tmp = "";
            for(int i=0;i<n;i++) {
                tmp += nums[i];
            }
            StringBuffer s = new StringBuffer(tmp);
            String r = "";
            while(k>0&&!s.toString().equals("")) {
                // 計算 (n-1)的排列個數cnt
                int cnt = 1, i = s.length()-1;
                while(i > 1) {
                    cnt*=i;
                    i-=1;
                }
                int pos = (k-1)/cnt;
                r += s.charAt(pos);
                s = s.deleteCharAt(pos);
                k -= pos * cnt;
            }
            return r;
        }