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, ifS=[1,2,3], A solution is:

[[3], [1], [2], [, 3], [], [], [], []
Thought: Read sequentially. Take each subset in the front and put the number of positions behind it as a new subset.

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, ifS=[1,2,2], A solution is:

[2], [1], [, 2], [], [], []
Thought: after sorting, follow the 1 method. However, if the preceding number is the same as the current number, read only each subset containing the preceding number and put itself behind it as a new subset.

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;    }};

 

42. subsets & subsets II