lintcode:Subsets II

•  Description
•  Notes
•  Testcase
•  Judge

Given a list of numbers that may has duplicate numbers, return all possible subsets

Have you met this question in a real interview? 

Yes

 Notice

• Each element in a subset must be in non-descending order.
• The ordering between two subsets is free.
• The solution set must not contain duplicate subsets.

Example

If S = [1,2,2], a solution is:

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

Challenge

Can you do it in both recursively and iteratively?

class Solution {
    
private:
    void subsetsWithDupHelper(const vector<int> &S, int curIdx, vector<int> &curVtr, vector<vector<int>> &retVtr, bool isPreChoosen) {
    
        // 在前面一个数和当前的数相同的情况下
        // 前选，当前可以选和不选  
        // 前不选， 当前不能选，只能不选
        
        if (curIdx == S.size())
        {
            retVtr.push_back(curVtr);
            return;
        }
        
        subsetsWithDupHelper(S, curIdx+1, curVtr, retVtr, false);
   
        if (curIdx > 0 && S[curIdx] == S[curIdx-1] && !isPreChoosen)
            return;
   
        curVtr.push_back(S[curIdx]);
        subsetsWithDupHelper(S, curIdx+1, curVtr, retVtr, true);
        curVtr.pop_back();
    }
    
public:
    /**
     * @param S: A set of numbers.
     * @return: A list of lists. All valid subsets.
     */
    vector<vector<int> > subsetsWithDup(vector<int> &S) {
        // write your code here
        
        vector<vector<int>> retVtr;
        if (S.size() == 0)
            return retVtr;
            
        sort(S.begin(), S.end()); //sort是关键之一
        vector<int> curVtr;
        
        subsetsWithDupHelper(S, 0, curVtr, retVtr, true);
        
        return retVtr;
    }
};