Permutation Sequence（超时，排列问题）

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.

Submission Result: Time Limit Exceede

Last executed input:

9, 94626

我的超时代码：

class Solution {
private:
    vector<vector<int>> res;
    int my_n;
    int my_k;
    bool overFlag;
public:
    void swap(vector<int> &a,int i,int j){
        int temp=a[i];
        a[i]=a[j];
        a[j]=temp;
    }
    void dfs(int dep,vector<int> temp,vector<int> v)
    {
        if(dep==my_n){
            res.push_back(temp);
            if(res.size()==my_k)
                overFlag=false;
            return;
        }
        for (int i=dep;i<my_n&&overFlag;++i)
        {
            swap(v,dep,i);
            temp.push_back(v[dep]);
            dfs(dep+,temp,v);
            temp.pop_back();
            swap(v,dep,i);
        }
    }
    string getPermutation(int n, int k) {
        overFlag=true;

        string resStr("");
        vector<int> v;
        my_n=n;
        my_k=k;
        for (int i=;i<=n;++i)
        {
            v.push_back(i);
        }
        vector<int> temp;
        dfs(,temp,v);

        vector<int> resV=res[k-];
        for (int i=;i<resV.size();++i)
        {
            resStr.push_back(resV[i]+'');
        }
        return resStr;
    }
};