求两个集合差的几种常见算法

有两个数组：

String[] arr01={"Andy","Bill","Cindy","Douglas","Felex","Green"};
String[] arr02={"Andy","Bill","Felex","Green","Gates"};
求存在于arr01而不存在于arr02的元素的集合？

 

最容易想到的解法-双重循环

import java.util.ArrayList;
import java.util.List;

/**
 * 利用双重循环实现的筛选
 */
public class DoubleCycling{
    public static void main(String[] args){
        String[] arr01={"Andy","Bill","Cindy","Douglas","Felex","Green"};
        String[] arr02={"Andy","Bill","Felex","Green","Gates"};

        // 筛选过程，注意其中异常的用途
        List<String> ls=new ArrayList<String>();
        for(String str:arr01){
            try{
                ls.add(getNotExistStr(str,arr02));
            }
            catch(Exception ex){
                continue;
            }
        }

        // 取得结果
        Object[] arr03=ls.toArray();
        for(Object str:arr03){
            System.out.println(str);
        }
    }

    /**
     * 查找数组Arr中是否包含str，若包含抛出异常，否则将str返回
     * @param str
     * @param arr
     * @return
     * @throws Exception
     */
    public static String getNotExistStr(String str,String[] arr) throws Exception{
        for(String temp:arr){
            if(temp.equals(str)){
                throw new Exception("");
            }
        }

        return str;
    }
}

 

 

速度较高的解法-利用哈希表

 

import java.util.ArrayList;
import java.util.Collection;
import java.util.Hashtable;
import java.util.List;
import java.util.Map;

/**
 * 利用哈希表进行筛选
 */
public class HashtableFilter{
    public static void main(String[] args){
        String[] arr01={"Andy","Bill","Cindy","Douglas","Felex","Green"};
        String[] arr02={"Andy","Bill","Felex","Green","Gates"};


        Map<String,String> ht=new Hashtable<String,String>();

        // 將arr02所有元素放入ht
        for(String str:arr02){
            ht.put(str, str);
        }

        // 取得在ht中不存在的arr01中的元素
        List<String> ls=new ArrayList<String>();
        for(String str:arr01){
            if(ht.containsKey(str)==false){
                ls.add(str);
            }
        }

        // 取得结果
        Object[] arr03=ls.toArray();
        for(Object str:arr03){
            System.out.println(str);
        }
    }
}

 

 

最方便的解法-利用工具类

 

import java.util.ArrayList;
import java.util.List;

/**
 * 使用工具类的筛选去除
 */
public class Tool{
    public static void main(String[] args){
        String[] arr01={"Andy","Bill","Cindy","Douglas","Felex","Green"};
        String[] arr02={"Andy","Bill","Felex","Green","Gates"};

        // 直接转的话，生成的List不支持removeAll
        List<String> ls01=new ArrayList<String>();
        for(String str:arr01){
            ls01.add(str);
        }

        // 同上
        List<String> ls02=new ArrayList<String>();
        for(String str:arr02){
            ls02.add(str);
        }

        // 去除arr01中存在于arr02中的元素
        ls01.removeAll(ls02);

        // 取得结果
        Object[] arr03=ls01.toArray();
        for(Object str:arr03){
            System.out.println(str);
        }
    }
}

 

利用二叉树的解法 

 

import java.util.ArrayList;
import java.util.List;

/**
 * 使用二叉樹的筛选去除
 */
public class Test{
    public static void main(String[] args){
        String[] arr01={"Andy","Bill","Cindy","Douglas","Felex","Green"};
        String[] arr02={"Andy","Bill","Felex","Green","Gates"};
font-size: 1em; margin-top: 0px; margin-righ

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JAVA布局管理器的（转） | (转)Math.Round() -- c# 与 java的区别

• 2010-10-21 10:34
• 浏览 575
• 评论(0)
• 分类:编程语言
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