Leetcode | | 78, subsets

Problem:

Given a set of distinct integers, S, return all possible subsets.

Note:

• Elements in a subset must is 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], [1,3], [+]  ,  [2,3], [up  ],  []]

Hide TagsArray backtracking Bit manipulationTest instructions: Given an array, find all its subsets, including the empty set

Thinking:

(1) 77 has already used the DFS to find the number of sets of all subsets of K, where the value range k is 1~n,n array length

Specific reference: http://blog.csdn.net/hustyangju/article/details/44974825

Code

Class Solution {private:    vector<vector<int> > ret;    Vector<int> tmp;public:    vector<vector<int> > Subsets (vector<int> &S) {    Ret.clear ();    unsigned int n=s.size ();    if (n==0)        return ret;    Sort (S.begin (), S.end ());    Tmp.clear ();    Ret.push_back (TMP);    for (int k=1;k<=n;k++)    {        tmp.resize (k);        DFS (0,n,s,k,0);    }    return ret;    } Protected:    void dfs (int dep, int n, vector<int> &s,int k,int start)    {       if (dep==k)       {           Ret.push_back (TMP);           return;       }       for (int i=start;i<n;i++)       {           tmp[dep]=s[i];           DFS (dep+1,n,s,k,i+1);}}    ;

Another Dfs method:

Class Solution {private:    vector<vector<int> > ret;public:    void dfs (int dep, int maxdep, vector< Int> &num, vector<int> A, int start)    {        ret.push_back (a);                if (dep = = MAXDEP)            return;                    for (int i = start; I < num.size (); i++)        {            vector<int> B (a);            B.push_back (Num[i]);            DFS (DEP + 1, MAXDEP, num, B, i + 1);        }    }        vector<vector<int> > Subsets (vector<int> &s) {        sort (s.begin (), S.end ());        Ret.clear ();        Vector<int> A;        DFS (0, s.size (), S, a, 0);                return ret;    }};

Leetcode | | 78, subsets