识别HashSet中的重复项

时间:2014-01-20 15:03:45

标签: java object add hashset

所以,我有这个用Java编写的代码:

import java.util.HashSet;

class Interval{
  long from;
  long to;

  public Interval(long from, long to) {
    this.from = from;
    this.to = to;
  }

  public boolean equals(Interval other) {

    return from == other.from && to == other.to;
  }


 }

public class Test {

   public static void main(String[] args) {

       HashSet<Interval> mySet  = new HashSet<Interval>();

       mySet.add(new Interval(1,2));
       mySet.add(new Interval(1,2));

       for(Interval in : mySet) {
        System.out.println(in.from + " " + in.to);
       }
   }

 }

问题是该集合无法识别已存在从1到2的间隔。我将函数定义为等于,但它仍然不起作用。我尝试实现Comparable接口并重载compareTo函数,但同样没有。有人能告诉我怎样才能解决这个问题?

谢谢!

2 个答案:

答案 0 :(得分:2)

您需要从equals覆盖java.lang.Object

你没有,因为你不接受Object作为参数。

public boolean equals(Object obj) {
    if (obj == null)
        return false;
    else if (this.getClass() != obj.getClass())
        return false;
    else {
        Interval other = (Interval) obj;
        return from == other.from && to == other.to;
    }
}

对于hashCode,您可以这样做。

public int hashCode() {
    return new Long(this.from).hashCode();
}

总的来说,你得到了这段代码。

import java.util.HashSet;

class Interval {
    long from;
    long to;

    public Interval(long from, long to) {
        this.from = from;
        this.to = to;
    }

    public boolean equals(Object obj) {
        if (obj == null)
            return false;
        else if (this.getClass() != obj.getClass())
            return false;
        else {
            Interval other = (Interval) obj;
            return from == other.from && to == other.to;
        }
    }

    public int hashCode() {
        return new Long(this.from).hashCode();
    }
}

public class Test003 {

    public static void main(String[] args) {

        HashSet<Interval> mySet = new HashSet<Interval>();

        mySet.add(new Interval(1, 2));
        mySet.add(new Interval2(1, 2));

        for (Interval in : mySet) {
            System.out.println(in.from + " " + in.to);
        }
    }

}

答案 1 :(得分:1)

使用如下所示的equals和hashCode方法,它可以正常工作

@Override
public int hashCode() {
    final int prime = 31;
    int result = 1;
    result = prime * result + (int) (from ^ from >>> 32);
    result = prime * result + (int) (to ^ to >>> 32);
    return result;
}

@Override
public boolean equals(Object obj) {
    if (this == obj) {
        return true;
    }
    if (obj == null) {
        return false;
    }
    if (getClass() != obj.getClass()) {
        return false;
    }
    Interval other = (Interval) obj;
    if (from != other.from) {
        return false;
    }
    if (to != other.to) {
        return false;
    }
    return true;
}