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.

[Thoughts]

两个解法。

第一，DFS

递归遍历所有可能，然后累加计算，直至到K为止。

第二，数学解法。

假设有n个元素，第K个permutation是

a1, a2, a3, .....   ..., an

那么a1是哪一个数字呢？

那么这里，我们把a1去掉，那么剩下的permutation为

a2, a3, .... .... an, 共计n-1个元素。 n-1个元素共有(n-1)!组排列，那么这里就可以知道

设变量K1 = K

a1 = K1 / (n-1)!

同理，a2的值可以推导为

a2 = K2 / (n-2)!

K2 = K1 % (n-1)!

 .......

a(n-1) = K(n-1) / 1!

K(n-1) = K(n-2) /2!

an = K(n-1)

实现如下：

1:       string getPermutation(int n, int k) {  2:            vector
    
      nums(n);  3:            int permCount =1;  4:            for(int i =0; i< n; i++)  5:            {  6:                 nums[i] = i+1;  7:                 permCount *= (i+1);        8:            }  9:            // change K from (1,n) to (0, n-1) to accord to index  10:            k--;  11:            string targetNum;  12:            for(int i =0; i< n; i++)  13:            {    14:                 permCount = permCount/ (n-i);  15:                 int choosed = k / permCount;  16:                 targetNum.push_back(nums[choosed] + '0');  17:                 //restruct nums since one num has been picked  18:                 for(int j =choosed; j< n-i; j++)  19:                 {  20:                      nums[j]=nums[j+1];  21:                 }  22:                 k = k%permCount;  23:            }  24:            return targetNum;  25:       }