1、全排列 

class Solution {
public:
    vector<vector<int>> ret;
    vector<int> path;
    bool check[7];

    void dfs(vector<int>& nums)
    {
        if(nums.size() == path.size()) 
        {
            ret.push_back(path);
            return;
        }
        for(int i = 0;i < nums.size();i++){
            if(check[i] == false)
            {
                path.push_back(nums[i]);
                check[i] = true;
                dfs(nums);
                //回溯
                path.pop_back();
                check[i] = false;
            }
        }
    }

    vector<vector<int>> permute(vector<int>& nums) {
        dfs(nums);
        return ret;
    }
};

 2、子集

解法一：根据每个元素选/不选来思考：  

解法二：根据元素个数来思考：

class Solution {
public:
    vector<int> path;
    vector<vector<int>> ret;
    vector<vector<int>> subsets(vector<int>& nums) {
        dfs(nums,0);
        return ret;
    }
    void dfs(vector<int>& nums,int pos)
    {
        ret.push_back(path);
        for(int i = pos;i < nums.size();i++){
            path.push_back(nums[i]);
            dfs(nums,i+1);
            path.pop_back();
        }
    }
};