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.

代码：

    string getPermutation(int n, int k) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        int fac[10];
        fac[0] = 1;
        for(int i = 1; i <= 9; i++){
            fac[i] = fac[i-1]*i;
        }
        string result = "";
        vector<int> nums;
        for(int i = 0; i < n; i++)
            nums.push_back(i+1);
        for(int i = n; i >= 1; i--){
            if(k <= fac[i-1]){
                result += (nums[0]+'0');
                nums.erase(nums.begin());
            }
            else if(k == fac[i]){
                for(vector<int>::iterator it = nums.end()-1; it >= nums.begin(); it--){
                    result += (*it+'0');
                }
                break;
            }
            else{
                int j = (k-1)/fac[i-1]+1;
                k = k - (j-1)*fac[i-1];
                vector<int>::iterator it = nums.begin();
                while(j > 1){
                    it++;
                    j--;
                }
                result += (*it+'0');
                nums.erase(it);
            }
        }
        return result;
    }