1.全排列

全排列

class Solution {vector<vector<int>> ret;vector<int> path;bool check[7];//标记nums数组某个下标是否已访问，剪枝使用
public:vector<vector<int>> permute(vector<int>& nums) {//递归+回溯+剪枝dfs(nums);return ret;}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--"恢复现场"path.pop_back();check[i] = false;}}}
};

2.子集

子集

class Solution {vector<vector<int>> ret;vector<int> path;
public:vector<vector<int>> subsets(vector<int>& nums) {dfs(nums,0);return ret;}// //法一：// void dfs(vector<int>& nums,int pos)// {//     if(pos == nums.size())//     {//         ret.push_back(path);//         return;//     }//     //选//     path.push_back(nums[pos]);//     dfs(nums,pos+1);//     path.pop_back();//     //不选//     dfs(nums,pos+1);// }//法二：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();}}
};