【LeetCode】78. Subsets (2 solutions)

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],
  []
]

解法一：

遍历S.size()位数的所有二进制数，1代表对应位置的元素在集合中，0代表不在。

一共2^n种情况。

详细步骤参照Subsets II

class Solution {
public:
    vector<vector<int> > subsets(vector<int> &S) {
        vector<vector<int> > result;
        int size = S.size();
        for(int i = ; i < pow(2.0, size); i ++)
        {//2^size subsets
            vector<int> cur;
            int tag = i;
            for(int j = size-; j >= ; j --)
            {//for each subset, the binary presentation has size digits
                if(tag% == )
                    cur.push_back(S[j]);
                tag >>= ;
                if(!tag)
                    break;
            }
            sort(cur.begin(), cur.end());
            result.push_back(cur);
        }
        return result;
    }
};

解法二：

遍历所有元素，记当前元素为S[i]

遍历当前所有获得的子集，记为ret[j]

将S[i]加入ret[j]，即构成了一个新子集。

详细步骤参照Subsets II

class Solution {
public:
    vector<vector<int> > subsets(vector<int> &S) {
        sort(S.begin(), S.end());
        vector<vector<int> > ret;
        vector<int> empty;
        ret.push_back(empty);
        for(int i = ; i < S.size(); i ++)
        {
            int size = ret.size();
            for(int j = ; j < size; j ++)
            {
                vector<int> newset = ret[j];
                newset.push_back(S[i]);
                ret.push_back(newset);
            }
        }
        return ret;
    }
};