Subsets

来源：互联网发布：上海市行知实验中学编辑：程序博客网时间：2024/06/16 05:01

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],  []]

public class Solution {    public List<List<Integer>> subsets(int[] S) {        int len=S.length;        ArrayList<ArrayList<Integer>> res=new ArrayList< ArrayList<Integer> >();        //if(S==null || S.length<=0) return res;        int x=1<<len;        for(int i=0;i<x;i++){                    res.add(getSubset(S,i));        }        Comparator<ArrayList<Integer>> comparator=new Comparator<ArrayList<Integer>>() {@Overridepublic int compare(ArrayList<Integer> a1,ArrayList<Integer> a2){    if(a1.size()>a2.size()){    return 1;    }    else if(a1.size()==a2.size()){    int  sum1=0,sum2=0;    for(Integer integer :a1){    sum1+=integer;    }    for(Integer integer :a2){    sum2+=integer;    }    if(sum1>sum2) return 1;    else return 0;    }    else return 0;    }};        Collections.sort(res, comparator);        ArrayList<List<Integer>> result=new ArrayList<List<Integer>>();        result.addAll(res);        return result;    }    public ArrayList<Integer> getSubset(int []S,int x){        ArrayList<Integer> a=new ArrayList<Integer>();        for(int i=S.length-1;i>=0;i--){            if((x&1)==1){                a.add(S[i]);            }            x>>=1;        }        //Arrays.sort(a);        Collections.sort(a);        return a;    }}

我的思路：对子集的结果要有序，只好进行排序，最后就写成这鸟样了。题目里给的解例子，好像是不合题意的。虽然AC，但是代码确实是过于冗长了。

C++版是这样的。

class Solution {public:    vector<vector<int> > subsets(vector<int> &S) {        sort(S.begin(), S.end());        vector<vector<int>> res;        vector<int> d;        int curr_size = 0;        res.push_back(vector<int>{});        for (int &e : S) {            curr_size = res.size();            for(int i = 0; i < curr_size; ++i) {                d = res[i];                d.push_back(e);                res.push_back(d);            }        }        return res;    }};

高手是这样解的：

def subsets(self, S):    res = [[]]    S.sort()    for e in S:        res = res+[l+[e] for l in res]    return res

这就是差距啊，所以还得努力学习啊。

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