算法描述：

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"

解題思路：理解排列與階乘的關系。n個數的排列方式有 n! 種，而n-1個數的排列方式有 (n-1)! 種。所以通過 k 除以 (n-1)! 得到對應位置的索引值。註意去除用過的數。

    string getPermutation(int n, int k) {
        vector<int> f(n+1
 
,0);
        f[0]=1;
        for(int i=1; i<=n; i++){
            f[i] = f[i-1]*i;
        }
        vector<int> num;
        for(int i =1; i<=n; i++){
            num.push_back(i);
        }
        k = k-1;
        string results = "";
        for(int i=n; i> 0; i--){
            int index = k / f[i-1
 
];
            k = k % f[i-1];
            results.push_back(num[index]+‘0‘);
            num.erase(num.begin()+index);
        }
        return results;
    }

LeetCode-60-Permuataion Sequence