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输入：nums = [1,2,3]
输出：[[],[1],[2],[1,2],[3],[1,3],[2,3],[1,2,3]]

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输入：nums = [0]
输出：[[],[0]]

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// @lc code=start
impl Solution {
    /// ## 解题思路
    /// - 回溯法
    pub fn subsets(nums: Vec<i32>) -> Vec<Vec<i32>> {
        // 遍历
        fn dfs(nums: &Vec<i32>, from: usize, tmp: &mut Vec<i32>, res: &mut Vec<Vec<i32>>) {
            // 终止条件
            if from > nums.len() {
                return;
            }
	    res.push(tmp.clone());
            // 遍历下一个
            for i in from..nums.len() {
                tmp.push(nums[i]);
                dfs(nums, i + 1, tmp, res);
                tmp.pop();
            }
        }

        let mut tmp = Vec::new();
        let mut res = Vec::new();
        dfs(&nums, 0, &mut tmp, &mut res);

        res
    }
}
// @lc code=end

struct Solution;

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class Solution {
    vector<vector<int>> result;
public:
    /*
    * ## 解题思路
    * - 深度遍历
    */
    vector<vector<int>> subsets(vector<int>& nums) {
        vector<int> tmp;
        result.push_back(tmp);
        dfs(nums, tmp, 0);
        return result;
    }

    void dfs(vector<int> &nums, vector<int> &tmp, int s) {
        if(s >= nums.size()) {
            return;
        }
        for(int i=s; i<nums.size(); i++ ) {
            tmp.push_back(nums[i]);
            result.push_back(tmp);
            dfs(nums, tmp, i+1);
            tmp.pop_back();
        }
    }
};