leetcode 78

DFS

只需要分nums[i]在不在subset中，DFS调用两次，一次将nums[i]推进去，一次nums[i]则不在结果中

void DFS(vector<vector<int>>& res, vector<int>& out, vector<int>& nums,int level) {
	if (level == nums.size()) {
		res.push_back(out);
		return;
	}
	else
	{
		DFS(res, out, nums, level+1);
		out.push_back(nums[level]);
		DFS(res, out, nums, level+1);
		out.pop_back();
	}
}
vector<vector<int>> subsets(vector<int>& nums) {
    vector<vector<int>> res;
	vector<int> out;
	DFS(res, out, nums,0);
	return res;
    }