好吧，我正在創建一個小游戲，我必須從數組列表中檢索獲勝者。private static ArrayList<Person> setWinner(ArrayList<Person> pers){ for(int i = 0; i < pers.size(); i++) { if(pers.get(i).getPoints() //Stuck here... } return pers; }如何查看哪些玩家得分最高？
6 回答
largeQ
TA貢獻2039條經驗 獲得超8個贊
像這樣的東西
if (players == null || players.size() == 0) {
throw new RuntimeException("You had no players!!");
}
Player winner = players.get(0);
for (Player p: players) {
if (p.getScore() > winner.getScore) {
winner = p;
}
}
winner; // here you got player with the most score.
慕尼黑的夜晚無繁華
TA貢獻1864條經驗 獲得超6個贊
由于出于某種原因您獲得了某種多個獲勝者，我認為獲勝者是比某些寶藏更多積分的玩家，并且可能只有固定數量的此類玩家（例如第 1、第 2、第 3 名）。
int winningScoreTreashold=5///somethis;
int winningPlayersCount=3;
List<Player> winners = players.stream()
.sorted(Comparator.comparing(Player::getScore))
.filter(p -> p.getScore() > winningScoreTreashold)
.limit(winningPlayersCount)
.collect(Collectors.toList());
winners//here you got winners;
刪除任何不符合您需求的條件。
慕慕森
TA貢獻1856條經驗 獲得超17個贊
像這樣的事情可以工作：
import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.List;
import java.util.stream.Collectors;
class Main {
public static void main(String[] args) {
List<Player> players = new ArrayList<>();
players.add(new Player("Adam", 1));
players.add(new Player("Bob"));
players.add(new Player("Cameron", 2));
players.add(new Player("Daniel", 2));
System.out.printf("Players: %s%n", players);
players.sort(Comparator.comparing(Player::getPoints));
System.out.printf("Sorted Players: %s%n", players);
List<Player> winners = getWinners(players);
System.out.printf("Winners: %s%n", winners);
}
public static List<Player> getWinners(List<Player> players) {
int maxPoints = Collections.max(players, Comparator.comparing(Player::getPoints)).getPoints();
return players.stream().filter(p -> p.getPoints() == maxPoints).collect(Collectors.toList());
}
}
假設你的Player類看起來像這樣：
class Player {
private String name;
private int points;
Player(String name, int points) {
this.name = name;
this.points = points;
}
Player(String name) {
this.name = name;
this.points = 0;
}
public void updatePoints(int newPoints) {
this.points = newPoints;
return;
}
public int getPoints() {
return this.points;
}
public String toString() {
return String.format("%s: %d", this.name, this.points);
}
}
輸出：
Players: [Adam: 1, Bob: 0, Cameron: 2, Daniel: 2]
Sorted Players: [Bob: 0, Adam: 1, Cameron: 2, Daniel: 2]
Winners: [Cameron: 2, Daniel: 2]
守著一只汪
TA貢獻1872條經驗 獲得超4個贊
使用以下解決方案怎么樣Streams：
private static List<Player> getWinners(List<Player> players) {
// Use TreeMap with reverse order comparator; the highest point will be the first
Supplier<Map<Integer, List<Player>>> mapFactory = () -> new TreeMap<>(Comparator.reverseOrder());
Map<Integer, List<Player>> map = Optional.of(players).orElse(Collections.emptyList()).stream()
.collect(Collectors.groupingBy(Player::getPoints, mapFactory, Collectors.toList()));
// map.keySet() is sorted desc order all points
return map.isEmpty() ? Collections.emptyList() : map.get(map.keySet().iterator().next());
}
MYYA
TA貢獻1868條經驗 獲得超4個贊
您需要首先對玩家數組進行排序。
返回與第一項的分數匹配的所有內容。
private static ArrayList<Player> getWinners(ArrayList<Player> players) {
if (players.isEmpty()) {
throw new RuntimeException("players list is empty");
}
// sort the array (desending)
Comparator<Player> comparator = (Player t, Player t1) -> t1.getPoints() - t.getPoints();
Collections.sort(players, comparator);
// filter the array
Predicate<Player> prdct = (Player t) -> t.score == players.get(0).getPoints();
List<Player> winners = players.stream().filter(prdct).collect(Collectors.toList());
return new ArrayList<>(winners);
}
希望這可以幫助！
翻閱古今
TA貢獻1780條經驗 獲得超5個贊
private static ArrayList<Player> getWinners(ArrayList<Player> players)
{
ArrayList<Player> l = new ArrayList<Player>();
int maxPoints = -1; //Assuming it is not possible to have negative points
for (int i = 0; i < players.size(); i++)
{
int points = players.get(i).getPoints();
if (points >= maxPoints) {
if (points > maxPoints) {
l.clear(); //Clear the return list, since a new "best score" was found
}
maxPoints = points;
l.add(players.get(i));
}
}
return l;
}
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