LeetCode-Subsets

作者：disappearedgod

文章出处：http://blog.csdn.net/disappearedgod/article/details/39757403

时间：2014-10-3

题目

Subsets

 Total Accepted: 23813 Total Submissions: 85835My Submissions

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],  []]

想法

这里就是简单模拟一个二进制的选择关系。

代码

public class Solution {    ArrayList<ArrayList<Integer>> retlist = new ArrayList<ArrayList<Integer>>();    public ArrayList<ArrayList<Integer>> subsets(int[] S) {        Arrays.sort(S);        int N = 1;        for(int i = 0; i < S.length; ++i)            N = N << 1;        if(S.length == 0)            return retlist;        for(int i = 0; i < N; i++){// retlist num            retlist.add(new ArrayList<Integer>());            for(int res = i,count = 0; res >0 && count < S.length; res = res >> 1,count++)                if(res % 2 == 1)                    retlist.get(i).add(S[count]);                    }        return retlist;    }}

结果

My Submissions for Subsets

Submit TimeStatusRun TimeLanguage4 minutes agoAccepted424 msjava

函数递归

函数递归思路参考了C++代码，但是这里要注意：

Java拷贝时候，list要用clone，否则传递为对象，所以会出现输出重复并不全的现象。

这里体现一个选择与不选择之间的区别。

代码

public class Solution {    ArrayList<ArrayList<Integer>> retlist = new ArrayList<ArrayList<Integer>>();    public ArrayList<ArrayList<Integer>> subsets(int[] S) {                ArrayList<Integer> list = new ArrayList<Integer>();        Arrays.sort(S);        sub(S, 0, new ArrayList<Integer>());        return retlist;            }    void sub(int[] S, int index, ArrayList<Integer> list){        if(index == S.length){            retlist.add(list);            return;        }                               //donot chose S[index]        sub(S, index+1, (ArrayList<Integer>)list.clone());                 //chose S[index]        list.add(S[index]);        sub(S, index+1, (ArrayList<Integer>)list.clone());                list.remove(list.size()-1);    }}

这里用了一个全集合的概念：

这个概念很像DFS

ArrayList<ArrayList<Integer>> retlist = new ArrayList<ArrayList<Integer>>();    public ArrayList<ArrayList<Integer>> subsets(int[] S) {        ArrayList<Integer> list = new ArrayList<Integer>();        Arrays.sort(S);        //sub(S, 0, new ArrayList<Integer>());        retlist.add(list);        combine( list, S, 0);        return retlist;    }        public void combine( ArrayList<Integer> list, int[] S, int index){        for(int i = index; i < S.length; ++i ){            list.add(S[i]);            retlist.add(new ArrayList<Integer>(list));//避免了clone            combine(list, S, i+1);            list.remove(list.size()-1);        }    }

class Solution {public:    void sub(vector<int> s, int index, vector<int> &path, vector<vector<int>> &solution)    {if(index == s.size()){solution.push_back(path);return;}sub(s,index+1,path,solution);path.push_back(s[index]);sub(s,index+1,path,solution);path.pop_back();}vector<vector<int>> subsets(vector<int> &S) {        // Note: The Solution object is instantiated only once.        vector<vector<int>> solution;vector<int> path;        sort(S.begin(),S.end());sub(S,0,path,solution);return solution;    }};

结果

My Submissions for Subsets

Submit TimeStatusRun TimeLanguage3 minutes agoAccepted412 msjava

My Submissions for Subsets

Submit TimeStatusRun TimeLanguage0 minutes agoAccepted40 mscpp

返回

 LeetCode Solution(持续更新，java>c++)

一些好的博客

         Leetcode Subset I & II         

子集系列(一) 传统subset 问题，例 [LeetCode] Subset, Subset II, Bloomberg 的一道面试题