OOP，实现对象的条件顺序输出，年龄或者姓名字符串长度

最新推荐文章于 2025-11-23 21:07:11 发布

原创最新推荐文章于 2025-11-23 21:07:11 发布 · 101 阅读

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#java #算法 #开发语言

package com.OOPtest;

public class test {
    public static void main(String[] args) {
        person p[] = new person[3];
        p[0] = new person("smith",20,"materials");
        p[1] = new person("jack",23,"com");
        p[2] = new person("An",22,"codo");

        for (int i = 0; i < p.length-1 ; i++) {
            for (int j = 0; j < p.length - 1 - i; j++) {

                if(p[j].getName().length() < p[j+1].getName().length() ) {
                    person p1 = p[j];
                    p[j] = p[j+1];
                    p[j+1] = p1;
                }

            }
        }

        for (int i = 0; i < p.length; i++) {
            System.out.println(p[i].toString());

        }

    }


}

class person{
    private String name;
    private int age;
    private String job;

    public person() {
    }

    public person(String name, int age, String job) {
        this.name = name;
        this.age = age;
        this.job = job;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public int getAge() {
        return age;
    }

    public void setAge(int age) {
        this.age = age;
    }

    public String getJob() {
        return job;
    }

    public void setJob(String job) {
        this.job = job;
    }

    @Override
    public String toString() {
        return "person{" +
                "name='" + name + '\'' +
                ", age=" + age +
                ", job='" + job + '\'' +
                '}';
    }



}