题目来源：https://leetcode.cn/problems/subsets/description/

 C++题解1：递归回溯法。由于是求子集，所以根据nums.size()遍历每个子集的长度，并进行回溯。

class Solution {
public:
    vector<vector<int>> res;
    vector<int> sub;
    void backtracking(vector<int> nums, int ind, int num) {
        if(sub.size() == num) {
            res.push_back(sub);
            return;
        }
        for(int i = ind; i < nums.size(); i++) {
            sub.push_back(nums[i]);
            backtracking(nums, i + 1, num);
            sub.pop_back();
        }
    }
    vector<vector<int>> subsets(vector<int>& nums) {
        int len = nums.size();
        for(int j = 0; j <= len; j++) {backtracking(nums, 0, j);}
        return res;
    }
};

C++题解2（来源代码随想录）：不断地从自身开始收集子集，直到末尾。所以不用更新目标子集的长度。

class Solution {
private:
    vector<vector<int>> result;
    vector<int> path;
    void backtracking(vector<int>& nums, int startIndex) {
        result.push_back(path); // 收集子集，要放在终止添加的上面，否则会漏掉自己
        if (startIndex >= nums.size()) { // 终止条件可以不加
            return;
        }
        for (int i = startIndex; i < nums.size(); i++) {
            path.push_back(nums[i]);
            backtracking(nums, i + 1);
            path.pop_back();
        }
    }
public:
    vector<vector<int>> subsets(vector<int>& nums) {
        result.clear();
        path.clear();
        backtracking(nums, 0);
        return result;
    }
};