class Solution {
public:
    /*参数len是需要走的路径长度了，container存储每个序列，u记录当前走到的长度*/
    void dfs(vector<int>& nums, vector<vector<int>> &container, int path[], int u, bool* st, int len){

        if(u == len){
            vector<int> tmp;
            for(int i = 0; i < len; i++) tmp.push_back(path[i]);

            /*下面这步是判断container中是否有重复序列*/
            bool found = false;
            for(int i =  0; i < container.size(); i++) 
                if(container[i] == tmp) found = true;
            if(!found) container.push_back(tmp);

            return ;
        }

        for(int i = 0 ; i < nums.size(); i++) {
            if(!st[i]){
                st[i] = true;
                path[u] = nums[i];
                dfs(nums, container, path, u + 1, st, len);
                st[i] = false;
            }
        }


    }

    vector<vector<int>> permutation(vector<int>& nums) {
        int path[nums.size()]; //存储路径
        vector<vector<int>> ans; //存储答案
        bool st[nums.size()]; //用来记录nums中的下标为i的元素是否被访问过
        int l = nums.size();
        dfs(nums,  ans, path, 0, st, l);
        return ans;
    }
};