Description：

• Difficulty:Medium
• Total Accepted:204.6K
• Total Submissions:475.3K
• Contributor:LeetCode

Given a set of distinct integers, nums, return all possible subsets (the power set).

Note: The solution set must not contain duplicate subsets.

For example,
If nums = [1,2,3], a solution is:

[
  [3],
  [1],
  [2],
  [1,2,3],
  [1,3],
  [2,3],
  [1,2],
  []
]

Thought:

在原有的集合中对每集合中的每个元素都加入新元素得到子集，然后放入原有集合中。

如果没有元素就返回空集（没有元素的数组）。

Code:

class Solution {
public:
	vector<vector<int>> subsets(vector<int>& nums) {
		return helper(nums, nums.size() - 1);
	}
	vector<vector<int>> helper(vector<int>& nums, int index)
	{
		if (index == -1)
		{
			vector<vector<int>> res =  vector<vector<int>>();
			vector<int> elem =  vector<int>();
			res.push_back(elem);
			return res;
		}
		vector<vector<int>> res = helper(nums, index - 1);
		int size = res.size();
		for (int i = 0; i<size; i++)
		{
			vector<int> elem = vector<int>(res[i]);
			elem.push_back(nums[index]);
			res.push_back(elem);
		}
		return res;
	}
};