Subsets II

 

Given a collection of integers that might contain duplicates, S, return all possible subsets.

Note:

• Elements in a subset must be in non-descending order.
• The solution set must not contain duplicate subsets.

 

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

[

  [2],

  [1],

  [1,2,2],

  [2,2],

  [1,2],

  []

]

 

有图为证：

class Solution {

public:

    void subReII(vector<vector<int> >& re, vector<int> &s,int j)

	{

		if(s.size() <= j)

			return;

		int size_ = re.size();

		int k = j + 1;

		//get the 重复区间,k是下个不重复的位置。

		while(k < s.size() && s[k] == s[k - 1]) k++;

		for(int i = 0; i < size_; ++i)

		{

			int cur = j;

			vector<int> copy(re[i]);

			//将重复的元素从 1 个到所有依次加入进去。

			while(cur < k)

			{

				copy.push_back(s[j]);

			    re.push_back(copy);

				++cur;

			}

		}

		//skip the 重复区间，到下一个不是重复的位置，递归。

		subReII(re, s, k);

	}

	vector<vector<int> > subsetsWithDup(vector<int> &S) {

		vector<vector<int> > re;

		vector<int> sol;

		sort(S.begin(),S.end());

		re.push_back(sol);

		subReII(re, S, 0);

		return re;

	}

};