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;    }};