java8排序

參考:http://www.javashuo.com/article/p-ubzrivgu-t.html

1.準備

實體類html

import lombok.AllArgsConstructor;
import lombok.Data;


@Data
@AllArgsConstructor
public class UserObj {
    private String name;
    private Integer age;
    private Integer credits;
}

測試類java

package com.pu.helloworld;

import org.assertj.core.util.Lists;

import java.util.List;

public class TestJava8 {
    public static void main(String[] args) {
        List<UserObj> userObjs = Lists.newArrayList(
                new UserObj("aa",9,70),
                new UserObj("aa",2,50),
                new UserObj("aa",6,60),
                new UserObj("ee",3,80),
                new UserObj("dd",5,70),
                new UserObj("ee",6,99),
                new UserObj("ff",9,73)
        );

    }
}

2.按年齡排序

這裏使用lambda表達式來代替匿名內部類,而且使用list接口下的sort方法(java8新增長),再鏈式輸出測試

   userObjs.sort(((o1, o2) -> o1.getAge()-o2.getAge()));
   userObjs.forEach(userObj -> System.out.println(userObj));

固然還能夠經過方法引用進一步的簡化,這裏使用Comparator下的comparingInt進行排序,使用User::getAge得到年齡,默認從小到大正向排序spa

userObjs.sort(Comparator.comparingInt(UserObj::getAge));
userObjs.forEach(System.out::println);

輸出:code

UserObj(name=aa, age=2, credits=50)
UserObj(name=ee, age=3, credits=80)
UserObj(name=dd, age=5, credits=70)
UserObj(name=aa, age=6, credits=60)
UserObj(name=ee, age=6, credits=99)
UserObj(name=aa, age=9, credits=70)
UserObj(name=ff, age=9, credits=73)htm

3.按年齡倒序

在比較器後面增長reversed便可,鏈式調用是java8的風格之一,一樣是閱讀性,原先的匿名內部類方法不只閱讀困難,一個簡單的倒序也須要先去觀察o2-o1仍是o1-o2才能得出,而Java8的方法不只代碼簡潔,可讀性還很高,compare getAge讀出是經過年齡進行排序,reversed讀出是倒序。blog

userObjs.sort(Comparator.comparingInt(UserObj::getAge).reversed());
userObjs.forEach(System.out::println);

4.按姓名,年齡,積分排序

  • 在這裏咱們使用比較器的thenComparing實現鏈式調用
 userObjs.sort(Comparator.comparing(UserObj::getName)
                .thenComparing(UserObj::getAge)
                .thenComparing(UserObj::getCredits)
        );
        userObjs.forEach(System.out::println);

輸出:排序

UserObj(name=aa, age=2, credits=50)
UserObj(name=aa, age=6, credits=60)
UserObj(name=aa, age=9, credits=70)
UserObj(name=dd, age=5, credits=70)
UserObj(name=ee, age=3, credits=80)
UserObj(name=ee, age=6, credits=99)
UserObj(name=ff, age=9, credits=73)接口

按姓名升,年齡降,積分升排序get

 userObjs.sort(Comparator.comparing(UserObj::getName)
                .thenComparing(UserObj::getAge,((o1, o2) -> o2-o1))
                .thenComparing(UserObj::getCredits)
        );

實上 o2 - o1 這樣的代碼仍是有一些命令式的風格,即包含了具體的實現過程(o2 -o1這樣的代碼),thenComparaing方法能夠直接接受一個排序器,所以咱們只要直接將倒序的排序器當作參數傳入便可,代碼以下

        userObjs.sort(Comparator.comparing(UserObj::getName)
                .thenComparing(Comparator.comparing(UserObj::getAge).reversed())
                .thenComparing(UserObj::getCredits)
        );
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