Java 如何使用最小值获取列表中的所有玩家

我试图以最低的尝试回报我比赛的赢家。但我不确定如何在打成平局的情况下做到这一点

我的目标是获得优胜者

try(Scanner scan = new Scanner(new File("result.txt"))){
                while(scan.hasNext()){
                    String[] s = scan.nextLine().split(" ");
                    players.add(new Players(s[0],s[1],Integer.valueOf(s[2])));
                }
                Collections.sort(players, Comparator.comparing((players1) -> players1.getAttempts()));
                //test// System.out.println(players);
                //Collections.sort(players, (a,b)->a.getAttempts().compareTo(b.getAttempts()));


                System.out.println("The winner is: "+ players.get(0).getName() +", with "+players.get(0).getAttempts()+ " attempt(s)!");
            }
            result.close();
}

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感谢您的帮助

首先根据尝试次数按升序对列表排序

List<Players> sortedList = players.stream().sorted((a,b)->a.getAttempts().compareTo(b.getAttempts())).collect(Collectors.toList())

或者简单地流式处理按升序排序的集合对象

players.filter(i->i.getAttempts()==players.get(0).getAttempts()).forEach(winner->System.out.println("The winners are "+winner.getvalues));

---

以下是另一种方法：

//read file into stream, try-with-resources
 try (Stream<String> stream = Files.lines(Paths.get("result.txt"))) {
            stream.map(line -> line.split("\\s"))
                  .map(a -> new Players(a[0], a[1], Integer.parseInt(a[2])))
                  .collect(groupingBy(Players::getAttemps))
                  .entrySet().stream().min(Comparator.comparingInt(Map.Entry::getKey))
                  .map(Map.Entry::getValue)
                  .map(l -> l.size() > 1 ?
                            l.stream().map(Players::getName)
                             .collect(joining(", ","there is a tie -->", 
                                     "with "+ l.get(0).getAttemps()+"attempt(s)!")) :
                           "The winner is: " + l.get(0).getName() +
                                   "with "+ l.get(0).getAttemps() +"attempt(s)!")
                  .ifPresent(System.out::println);
} catch (IOException e) { e.printStackTrace(); }

---

在平局时需要做什么？显示第一名有平局，然后列出玩家的姓名，以及尝试在

新玩家（s[0]，s[1]，Integer.valueOf（s[2]）中哪些属性及其对应的类型

？另外，基于值，您当前没有进行比较。对于过滤器，它说找不到符号。您能解释一下吗？这次在@comp进行了一些更改。Collect说找不到符号。orry

Collect（）

小写字母非常简单，只需将这一行更改为

a.getAttempts（）.compareTo（b.getAttempts（））

像这样

//read file into stream, try-with-resources
 try (Stream<String> stream = Files.lines(Paths.get("result.txt"))) {
            stream.map(line -> line.split("\\s"))
                  .map(a -> new Players(a[0], a[1], Integer.parseInt(a[2])))
                  .collect(groupingBy(Players::getAttemps))
                  .entrySet().stream().min(Comparator.comparingInt(Map.Entry::getKey))
                  .map(Map.Entry::getValue)
                  .map(l -> l.size() > 1 ?
                            l.stream().map(Players::getName)
                             .collect(joining(", ","there is a tie -->", 
                                     "with "+ l.get(0).getAttemps()+"attempt(s)!")) :
                           "The winner is: " + l.get(0).getName() +
                                   "with "+ l.get(0).getAttemps() +"attempt(s)!")
                  .ifPresent(System.out::println);
} catch (IOException e) { e.printStackTrace(); }

import static java.util.stream.Collectors.*;
import java.util.stream.*;
import java.util.List;
import java.util.*;
import java.nio.file.Paths;
import java.nio.file.Files;
import java.io.*;