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();}}
};