1. 题意

给定一个不含有重复数字的数列，求所有的子集。

2. 题解

子集型回溯，可以直接用dfs进行搜索；也可以用二进制来进行枚举。

2.1 选或不选

class Solution {
public:
    void dfs(vector<vector<int>> &ans,vector<int> &tmp,
             vector<int> &nums, int depth) {
        if (depth == nums.size()) {
            ans.emplace_back(tmp );
            return;
        }

            dfs(ans, tmp, nums, depth + 1);
            tmp.push_back(nums[depth]);
            dfs(ans, tmp, nums, depth + 1);
            tmp.pop_back();
    }


    vector<vector<int>> subsets(vector<int>& nums) {
        
        vector<vector<int>> ans;
        vector<int> tmp;
        dfs( ans, tmp, nums, 0);
        return ans;
    }
};

2.2 选哪个

class Solution {
public:
    void dfs(vector<vector<int>> &ans,vector<int> &tmp,
             vector<int> &nums, int depth) {
        
        ans.emplace_back(tmp);
        for (int i = depth;i < nums.size(); i++) {
            tmp.push_back(nums[i]);
            dfs(ans, tmp, nums, i + 1);
            tmp.pop_back();
        }
    }


    vector<vector<int>> subsets(vector<int>& nums) {
        
        vector<vector<int>> ans;
        vector<int> tmp;
        dfs( ans, tmp, nums, 0);
        return ans;
    }
};

2.3 二进制枚举

class Solution {
public:
    vector<vector<int>> subsets(vector<int>& nums) {
        
        vector<vector<int>> ans;
        int sz = nums.size();

        for (int i = 0;i < (1 << sz); i++) {
            vector<int> tmp;
            for (int j = 0;j < sz;j++) {
                if (i & (1 << j))
                    tmp.push_back(nums[j]);
            }
            ans.emplace_back( tmp );
        }

        return ans;
    }
};

参考

0x3f题解