如何在Java </object>中按对象属性对arrayList <object>进行排序

时间:2013-01-25 10:15:25

标签: java sorting collections compare long-integer

我有类(JavaBean,如果你想这样称呼它)

class Tweet{

private millis; //number of millis since 1970
//other attributes and getters and setters, but i want to sort onlny by millis

public long getMillis() {
   return millis;
}

}

比较者应该看起来像这样:

class TweetComparator implements Comparator {
     @Override
     public int  compare(Tweet t1, Tweet t2) {
     //something
     //this doesn't work
     //return t2.getMillis().compareTo(t1.getMillis());
     return ??;//what should be here?
     }

    }

这将在程序中

List<Tweet> tweets = new ArrayList<Tweet>();
tweets.add(...); //just fill the list
//i need newest (with hightest millis value first) so I probably need to call reverse order
Collection.reverse(tweets)
Collection.sort(tweets, new TweetComparator());

我找到了一些参考文献herehere。但我不知道如何完成我的代码。

3 个答案:

答案 0 :(得分:11)

您的比较器应该与此类似

class TweetComparator implements Comparator<Tweet> {
    @Override
    public int compare(Tweet t1, Tweet t2) {
        return Long.compare(t1.getMillis(), t2.getMillis()); 
    }
}

请注意,static int Long.compare是自Java 7以来的

答案 1 :(得分:4)

  

比较方法返回:    第一个参数的负整数,零或正整数小于,等于,>>或大于第二个参数。

逻辑 -

if t1.millis > t2.millis 
   return -1;
else if t1.millis < t2.millis
   return +1;

代码 -

class TweetComparator implements Comparator<Tweet> {
     @Override
     public int  compare(Tweet t1, Tweet t2) {
        if(s1.i>s2.i)
            return -1;
        else if(s1.i<s2.i)
            return +1;
        return 0;
     }

 }

答案 2 :(得分:3)

试试这个:

@Override
     public int  compare(Tweet t1, Tweet t2) {

     return t1.getMillis().compareTo(t2.getMillis());

     }

如果要使用Long类的内置compareTo方法,请将mills变为long。 否则在比较方法中,将您的millis与t1和t2进行比较,如下所示。

long t1Val = t1.getMillis();
long t2Val = t2.getMillis();
return (t1Val<t2Val? -1 : (t1Val ==t2Val? 0 : 1));

(直接来自原版Long课程)