典型的遞迴狀態生成問題。類似於全排列的生成問題。

問題一：無重複元素集合的子集。Given a set of distinct integers, S, return all possible subsets.

思路：藉助一個now陣列儲存臨時的子集。不斷切換狀態進行深層遞迴，在遞迴最深處儲存生成的子集。

class Solution {
public:
    vector<vector<int> > subsets(vector<int> &S) {
        vector<vector<int> > result;
        if(S.empty()) return result;
        sort(S.begin(), S.end());
        vector<int> now;
        fun(S, result, now, 0);
        return result;
    }
    
    void fun(const vector<int> &s, vector<vector<int> > &result, vector<int> &now, int n)
    {
        if(n == s.size())
        {
            result.push_back(now); //遞迴最深處：所有元素都判定過取或不取
            return;
        }
        else
        {
            fun(s, result, now, n+1); //不取
            now.push_back(s[n]);
            fun(s, result, now, n+1); //取
            now.pop_back(); //回溯前一定要恢復回原樣
        }
        
    }
};
 

問題二：有重複元素集合的子集。Given a collection of integers that might contain duplicates, S, return all possible subsets.

class Solution {
public:
    vector<vector<int> > subsetsWithDup(vector<int> &S) {
        vector<vector<int> > result;
        vector<int> now;
        sort(S.begin(), S.end());//排序
        set<vector<int> > s;
        backtrack(s, now, S, 0);
        set<vector<int> >::iterator it = s.begin();
        for(;it != s.end();it++)
            result.push_back(*it);
        return result; 
    }
    
    void backtrack(set<vector<int> > &result, vector<int> &now, vector<int> &S, int idx)
    {
        if(idx == S.size())
        {
            result.insert(now);//set去重複
            return;
        }
        backtrack(result, now, S, idx+1);
        now.push_back(S[idx]);
        backtrack(result, now, S, idx+1);
        now.pop_back();
    }
};