115_leetcode_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 (ie, 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.

1:特殊情况的考虑；2:按照特定的规律从数组中取出相应的数字，放在字符串中，并删除数组中的相应数字；3:注意循环结束情况

    string getPermutation(int n , int k)
    {
        if(n <= 0 || k <= 0)
        {
            return "";
        }
        
        int count = 1;
        for(int i = 1; i <= n; i++)
        {
            count *= i;
        }
        
        if(k > count)
        {
            return "";
        }
        
        string result;
        vector<int> myVector;
        for(int i = 1; i <= n; i++)
        {
            myVector.push_back(i);
        }
        
        int index = n;
        while(index > 0)
        {

            count /= index;
            int curindex = (k - 1) / count;
            result.push_back(myVector[curindex] + '0');
            myVector.erase(myVector.begin() + curindex);
            k -= count * curindex;
            index = index - 1;
            
        }
    
        return result;
    }