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);
    }
};