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,
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>>() {
			
			@Override
			public 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

Subsets,布布扣,bubuko.com

Subsets

原文：http://blog.csdn.net/dutsoft/article/details/38341351