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();
}
}
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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