60. 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 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.
• Given k will be between 1 and n! inclusive.

Example 1:

Input: n = 3, k = 3
Output: "213"

Example 2:

Input: n = 4, k = 9
Output: "2314"

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 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.
• Given k will be between 1 and n! inclusive.

Example 1:

Input: n = 3, k = 3
Output: "213"

Example 2:

Input: n = 4, k = 9
Output: "2314"

把1~n个的数字按字典序排列，然后返回第k个排列。

不需要把所有排序都列出来，首先，筛选第一位，则后面的n-1位有(n-1)!的排列，则用看k是n-1的多少倍，则可以选出第一个数字。下一个数字则k - 【k / (n - 1)! 】* (n-1)！ 以此列推。

class Solution {
        public int  jiecheng (int k) {
            int rt = 1;
            for (int i = 1; i <= k; i++)
                rt *= i;
            return rt;
        }

        public String getPermutation(int n, int k) {
            String ans = "";
            //用一个数组储存已经选过的数字
            boolean []visited = new boolean[n];
            int count = 0;
            int num = n;
            n--;
            while (count < num) {
                int jk = jiecheng(n);
                //计算出，应该选择剩余数字的第几个。
                int select = ( k - 1) / jk;
                int index = 0;
                int  i = 0;
                //选出未未选出的第select个
                while (true) {
                    if (!visited[i]) {
                        if (index == select) {
                            ans += (i + 1);
                            visited[i] = true;
                            break;
                        }
                        index++;
                    }
                    i++;
                }
                //减去当前排序数量，进入下一位
                k = k - select * jk;
                count++;
                n--;
            }
            return ans;
        }
}