LeetCode题解：Subsets I and II

Subsets

Given a set of distinct integers, 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 = [1,2,3], a solution is:

[
  [3],
  [1],
  [2],
  [1,2,3],
  [1,3],
  [2,3],
  [1,2],
  []
]

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：
排列组合题。对n个元素的集合S，一共有2^n个子集。其中每个元素都有“取”和“不取”两种状态。这样自然而然的就想到了用位操作解决。考虑到解长度的指数增长，不需要使用bitset之类的，直接用size_t已经足够了。

对于第二题，需要对每个相同元素的个数进行计数，然后递归即可。
题解：

class Solution {
public:
    vector<vector<int> > subsets(vector<int> &S) {
        sort(begin(S), end(S)); // ensure non-descending results
        
        const size_t n = S.size();
        const size_t num_sets = pow(2, n);
        
        vector<vector<int>> ret(num_sets);
    
        for(size_t bitwise = 0; bitwise < num_sets; ++bitwise)
        {
            size_t bit_chk = 1;
            for(size_t bit = 0; bit < n; ++bit)
                if ((bit_chk << bit) & bitwise)
                    ret[bitwise].push_back(S[bit]);
        }
        
        return ret;
    }
};

class Solution {
public:
    vector<int> buffer;
    vector<vector<int>> result;
    vector<pair<int, int>> n_count;
    
    void build(vector<pair<int, int>>::iterator iter)
    {
        if (iter == end(n_count))
        {
            result.push_back(buffer);
            return;
        }
        
        for(int i = 0; i <= iter->second; ++i)
        {
            build(next(iter));
            
            buffer.push_back(iter->first);
        }
        
        for(int i = 0; i <= iter->second; ++i)
            buffer.pop_back();
    }
    
    vector<vector<int> > subsetsWithDup(vector<int> &S) {
        result.clear();
        if (S.empty())
            return result;
        
        n_count.clear();
        
        sort(begin(S), end(S));
        
        int v = S.front();
        int c = 0;
        for(auto& i : S)
        {
            if (v == i)
                ++c;
            else
            {
                n_count.push_back(make_pair(v, c));
                v = i;
                c = 1;
            }
        }
        n_count.push_back(make_pair(v, c));

        build(begin(n_count));
        return result;
    }
};