leetcode：回溯——permutation-sequence，

1. permutation-sequence 顺序排列第k个序列

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 k th permutation sequence.

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

第一位每个数字开头的序列都有（n-1）！个序列，因此n个数字所以共有n！个序列。以此类推，第二位每一个数开头都有（n-2）！个序列。

 

public class Solution {
    public String getPermutation(int n, int k) {
 
        // initialize all numbers
        ArrayList<Integer> numberList = new ArrayList<Integer>();
        for (int i = 1; i <= n; i++) {
            numberList.add(i);
        }
 
        // change k to be index
        k--;
 
        // set factorial of n
        int mod = 1;
        for (int i = 1; i <= n; i++) {
            mod = mod * i;
        }
 
        String result = "";
 
        // find sequence
        for (int i = 0; i < n; i++) {
            mod = mod / (n - i);
            // find the right number(curIndex) of
            int curIndex = k / mod;
            // update k
            k = k % mod;
 
            // get number according to curIndex
            result += numberList.get(curIndex);
            // remove from list
            numberList.remove(curIndex);
        }
 
        return result.toString();
    }
}