问题描述

我有那个结构的集合.我没有重复，但是当我打电话时:set.add(element) -> 并且已经有确切的元素我想替换旧的.

I have Set of that structure. I do not have duplicates but when I call: set.add(element) -> and there is already exact element I would like the old to be replaced.

import java.io.*;

public class WordInfo implements Serializable {
    File plik;
    Integer wystapienia;

    public WordInfo(File plik, Integer wystapienia) {
        this.plik = plik;
        this.wystapienia = wystapienia;
    }

    public String toString() {
    //  if (plik.getAbsolutePath().contains("src") && wystapienia != 0)
            return plik.getAbsolutePath() + "	WYSTAPIEN " + wystapienia;
    //  return "";
    }
    @Override
    public boolean equals(Object obj) {
        if(this == obj) return true;
        if(!(obj instanceof WordInfo)) return false;
        return this.plik.equals(((WordInfo) obj).plik);
    }

    @Override
    public int hashCode() {        
        return this.plik.hashCode();
    }
}

推荐答案

每次添加前做一个remove:

Do a remove before each add:

 someSet.remove(myObject);
 someSet.add(myObject);

remove 将删除任何等于 myObject 的对象.或者，您可以检查添加结果:

The remove will remove any object that is equal to myObject. Alternatively, you can check the add result:

 if(!someSet.add(myObject)) {
     someSet.remove(myObject);
     someSet.add(myObject);
 }

哪种方式更有效取决于您发生碰撞的频率.如果它们很少见，第二种形式通常只做一次操作，但当发生碰撞时，它会做三次.第一种形式总是做两个.

Which would be more efficient depends on how often you have collisions. If they are rare, the second form will usually do only one operation, but when there is a collision it does three. The first form always does two.

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