leetcode || 78、Subsets

problem：

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],
  []
]

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Array Backtracking Bit
Manipulation

题意：给定一个数组，求出其所有的子集，包括空集

thinking：

（1）77题中已经使用DFS求出数组的指定个数为K的所有子集，这里K的取值范围为1~n，n为数组长度

具体参考：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);
       }
    }
};

另一种DFS法：

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;
    }
};