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],
  []
]

思路
在上一题的基础上，因为数组元素可能有重复，所以在递归的时候要避免如果有几个相同的元素，在这几个元素中间选择时只能往上增加，不能跳跃的选择。因此增加第24行到第29行的判断。

   vector<vector<int> > result;
    vector<int> tmp;
    void initialize(vector<int> &S){
        tmp.clear();
        result.clear();
        int n = S.size();
        int i,j;
        int t;
        for(i = 0; i < n-1; i++){
            for(j = 0; j <= n-2-i; j++){
                if(S[j] > S[j+1]){
                    t = S[j];
                    S[j] = S[j+1];
                    S[j+1] = t;
                }
            }
        }
    }
    void getSubsets(vector<int> &S, int start, int end){
        if(end == start){
            result.push_back(tmp);
            return;
        }
        if(tmp.empty())
            getSubsets(S, start+1, end);
        else{
            if(S[start] > tmp[tmp.size()-1])
                getSubsets(S, start+1, end);
        }
        tmp.push_back(S[start]);
        getSubsets(S, start+1, end);
        tmp.pop_back();
    }
    vector<vector<int> > subsetsWithDup(vector<int> &S) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        int n = S.size();
        initialize(S);
        getSubsets(S, 0, n);
        return result;
    }