leetcode 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],
  []
]

 深搜即可

　如果想更加深入的理解集合的子集问题可以阅读《Efficiently Enumerating the Subsets of a Set》

typedef vector<vector<int> > VectorArray;
vector<int> solution;
VectorArray result;

void  dfs(int k, vector<int>& S){
    for(int i = k; i < S.size(); ++ i){
        solution.push_back(S[i]);
        result.push_back(solution);
        dfs(i+1,S);
        solution.pop_back();
    }
}
VectorArray subsets(vector<int>& S){
    sort(S.begin(),S.end());
    result.push_back(solution);
    dfs(0,S);
    return result;
}

 另一个比较简单的方法，根据集合生产规则，迭代生产，可以手动模拟一下{1，2，3}

最初子集为{{}}

添加一个元素后为{{}，{1}}，则在现有的集合上添加下一个元素，现有子集{}，{1}

添加元素2后为{2}，{1，2}，加入子集为{{}，{1}，{2}，{1，2}}，在现有的子集上添加集合

添加元素3后为{3}，{1，3}，{2，3}，{1，2，3}，加入子集后

得到

{{}，{1}，{2}，{3}，{1，2}，{1，3}，{2，3}，{1，2，3}}

vector<vector<int> > subsets(vector<int>& S){
    vector< vector<int> > result;
    vector<int> empty;
    result.push_back( empty );

    for (int i = 0; i < set.size(); i++)
    {
        vector< vector<int> > subsetTemp = result;

        for (int j = 0; j < subsetTemp.size(); j++)
            subsetTemp[j].push_back( set[i] );

        for (int j = 0; j < subsetTemp.size(); j++)
            result.push_back( subsetTemp[j] );
    }
    return result;
}

 还有一种方法是利用位实现集合，可以参考http://codeding.com/?article=12，

如果不能打开，可以参考下面一句话

There will be 2^N subsets for a given set, where N is the cardinality (number of items) of that set. For example, there will be 2⁵ = 32 subsets for the set {1, 2, 3, 4, 5}. The binary representation of each number 'i' in the range 0 to (2^N - 1) is used to find the corresponding i^th subset.

Each '1' in the binary representation indicates an item in that position. For example, the binary representation of number 5 is 00101 which in turn represents the subset {3, 5} of the set {1, 2, 3, 4, 5}.