HashSet判断重复项原理

Set集合的方法全部来自于继承Collection

先用hashCode计算存到哪个位置,为空直接保存再用equals判断是否重复不重复就串一个单向链表

先是hashCode再是equals这两个方法返回的结果都是相等,才判定为重复 Student实体类 基本配置

public class Student implements Comparable{
public String name;
public int age;
public char sex;

public Student(){}

public Student(String name,int age,char sex){
    this.age = age;
    this.name = name;
    this.sex = sex;
}

相同数据的字段的hashCode值都是一样的:都是从常量池取值重写hashCode本质:就是判断各个属性的hashCode值,是否相等 为什么乘以31:

31是一个素质数(只有1和它本身两个因数),可以减少散列冲突,尽量让计算结果不同

提高执行效率:31*i = (i << 5) -i:31本身就是2的4次方-1

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这是数学家们应该探讨的问题,我们只要知道这个事情就可以了

@Override
public int hashCode() {
    //return (this.sex + this.age + this.name).hashCode() * 31;
    
    return Objects.hash(name, age, sex);
}

@Override
public boolean equals(Object o) {
    if (this == o) return true;
    if (o == null || getClass() != o.getClass()) return false;

    if(o instanceof Student){
        Student student = (Student) o;
        return age == student.age && sex == student.sex && Objects.equals(name, student.name);
    }
    return  false;
}

package com.li.changGe.collections.setGather;

import com.li.changGe.pojo.Student;
import java.util.HashSet;
import java.util.Set;

public class HashSetDemo01 {
  public static void main(String[] args) {
    
    Set set = new HashSet();

    Student student = new Student("长歌",18,'女');
    Student student1 = new Student("世民",22,'男');
    Student student2 = new Student("则天",20,'女');

    set.add(student);
    set.add(student1);
    set.add(student2);
//------------------------------------------------
    
    set.add(new Student("长歌",18,'女'));

    
    System.out.println("长度:"+set.size()+"n"+set);

  }

}