题目传送地址:​​https://leetcode.cn/problems/permutation-sequence/​​​ 先给大家看一段用全排列的递归解法，但是不符合题目的性能要求。
代码：

public static String getPermutation(int n, int k) {
        String[] allArrange = getAllArrange(n);
        Arrays.sort(allArrange);
        return allArrange[k - 1];
    }

    /**
     * 全排列
     *
     * @param n
     * @return
     */
    public static String[] getAllArrange(int n) {
        if (n == 1) {
            String[] res = new String[1];
            res[0] = "1";
            return res;
        }
        List<String> list = new ArrayList<>();
        String[] allArrange = getAllArrange(n - 1);
        for (int i = 0; i < allArrange.length; i++) {
            String str = allArrange[i];
            for (int j = 0; j <= str.length(); j++) {
                StringBuilder stringBuilder = new StringBuilder(str);
                stringBuilder.insert(j, n);
                list.add(stringBuilder.toString());
            }
        }
        String[] res;
        res = list.toArray(new String[0]);
        return res;
    }

真正厉害的代码是下面这样

public static String getPermutation(int n, int k) {
        StringBuilder stringBuilder = new StringBuilder();
        while (n > 0) {
            stringBuilder.insert(0, n);
            n--;
        }
        //比如n=4,则result=1234
        return getResult(stringBuilder.toString(), k);
    }

    public static String getResult(String str, int k) {
        //处理边界情况
        if (str.length() == 1) {
            return str;
        }
        int index = 0;
        int factorial = doFactorial(str.length() - 1);
        //计算出第一个字符   //比如n=4,k=9的情况下，本题的正确答案应该是"2314"，  factorial=6,
        // 而9/6=1,经过下面的while循环以后，fisrtChar=2,  k=3
        while (k > factorial) {
            index++;
            k -= factorial;
        }
        if (k == 0 && index == 0) {
            return str;
        }
        char firstChar = str.charAt(index);
        String res = new StringBuilder(str).deleteCharAt(index).toString();
        String result = getResult(res, k);
        StringBuilder stringBuilder = new StringBuilder(result);
        stringBuilder.insert(0, firstChar);
        return stringBuilder.toString();
    }

    /**
     * 阶乘
     * @param m
     * @return
     */
    public static int doFactorial(int m) {
        if (m == 1) {
            return 1;
        }
        return doFactorial(m - 1) * m;
    }

运行效率：

Leetcode60. 排列序列_while循环