42. Subsets && Subsets 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],
  []
]

思想： 顺序读，取前面的每个子集，把该位置数放后面作为新的子集。

class Solution {
public:
    vector<vector<int> > subsets(vector<int> &S) {
        sort(S.begin(), S.end());
        vector<vector<int> > vec(1);
        for(size_t id = 0; id < S.size(); ++id) {
            int n = vec.size();
            while(n-- > 0) {
                vec.push_back(vec[n]);
                vec.back().push_back(S[id]);
            }
        }
        return vec;
    }
};

 

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],
  []
]
思想： 排序后，按照 1 的方法。但是若前面的数字与本数字相同，则只读取含有前面数字的每个子集，把自身放在后面作为一个新的子集。

class Solution {
public:
    vector<vector<int> > subsetsWithDup(vector<int> &S) {
        sort(S.begin(), S.end());
        vector<vector<int> > vec(1);
        size_t prePos, endTag;
        prePos = endTag = 0;
        for(size_t id = 0; id < S.size(); ++id) {
            if(id > 0 && S[id] != S[id-1]) endTag = 0;
            else endTag = prePos;
            size_t n = vec.size();
            prePos = n;
            while(n > endTag) {
                --n;
                vec.push_back(vec[n]);
                vec.back().push_back(S[id]);
            }
        }
        return vec;
    }
};