17-02Java语言基础（HashMap）

HashMap

把Student对象作为键进行存储：

Student.java：

package com.xiaoxin.bean;

public class Student {
     
	private String name;
	private int age;
	
	public Student(String name, int age) {
     
		super();
		this.name = name;
		this.age = age;
	}

	@Override
	public String toString() {
     
		return "Student [name=" + name + ", age=" + age + "]";
	}

	@Override
	public int hashCode() {
     
		final int prime = 31;
		int result = 1;
		result = prime * result + age;
		result = prime * result + ((name == null) ? 0 : name.hashCode());
		return result;
	}

	@Override
	public boolean equals(Object obj) {
     
		if (this == obj)
			return true;
		if (obj == null)
			return false;
		if (getClass() != obj.getClass())
			return false;
		Student other = (Student) obj;
		if (age != other.age)
			return false;
		if (name == null) {
     
			if (other.name != null)
				return false;
		} else if (!name.equals(other.name))
			return false;
		return true;
	}
	
}

Demo4_HashMap.java：

package com.xiaoxin.map;

import com.xiaoxin.bean.Student;
import java.util.HashMap;

public class Demo4_HashMap {
     
	
	//HashMap集合键是Student，值是String案例
	//键是学生对象
	//值是字符串对象，代表学生归属地
	public static void main(String[] args) {
     
		HashMap<Student, String> hm = new HashMap<>();
		hm.put(new Student("张三", 23), "北京");
		hm.put(new Student("张三", 23), "东北");
		hm.put(new Student("王五", 25), "东京");
		hm.put(new Student("赵六", 26), "武汉");
		
		System.out.println(hm);
	}
}

输出：

{Student [name=张三, age=23]=东北, Student [name=赵六, age=26]=武汉, Student [name=王五, age=25]=东京}

分析：

需要在Student类中重写hashCode()和equals()方法，才能在输入的时候判断Student键有没有重复