Subsets II

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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;}};

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