思路一：迭代法

代码

class Solution {
public:vector<vector<int>> subsets(vector<int>& nums) {vector<vector<int>> res;int n = nums.size();for(int i = 0; i < 1 << n; i++) // 1 << n = 2^n{vector<int> path;for(int j = 0; j < n; j++){if(i >> j & 1) path.push_back(nums[j]);}res.push_back(path);}return res;}
};

方法二:经典dfs

代码

class Solution {
public:vector<int> path;vector<vector<int>> res;vector<vector<int>> subsets(vector<int>& nums) {dfs(0,nums);return res;}void dfs(int count, vector<int>& nums){if(count == nums.size()){res.push_back(path);return;}path.push_back(nums[count]); //将该数据加进来的路径dfs(count + 1, nums);path.pop_back(); // 将该数据清除（不加进去）后的路径dfs(count + 1, nums);}
};