leetcode || 78、Subsets

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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;    }};

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