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我有一个即将举行的国际足联锦标赛，我编写了一个程序来打印出可能的比赛。问题是它没有按逻辑排序，这意味着一些玩家必须连续玩 5-6 场比赛，而另一些则必须等待 6 场比赛。我想得到以下结果：

player 1 - player 2
player 3 - player 4
player 5 - player 6
player 1 - player 3
player 2 - player 4

等等。这就是我目前所拥有的：

public class Fifa {

public static void main(String[] args) {
    String[] players= {"Jens", "Dane", "Keppens", "Roel", "John", "Onslo", "JonasDB", "Bellon", "Sander"};
    String[] players2 = {"Jens", "Dane", "Keppens", "Roel", "John", "Onslo", "JonasDB", "Bellon", "Sander"};


    Multimap<String, String> fixtures = LinkedHashMultimap.create();

    for(int i = 0; i < players.length; i++){
        for (int j = 0; j < players.length; j++){
            if(!players[i].equals(players2[j])) {
                if(!fixtures.containsKey(players2[j]))
                fixtures.put(players[i], players2[j]);
            }
        }
    }

    for(Map.Entry map : fixtures.entries()){
        String key = map.getKey().toString();
        Object value = map.getValue();
        System.out.println(key + " - " + value);
    }

但这是打印出来的：

Jens - Dane
Jens - Keppens
Jens - Roel
Jens - John
Jens - Onslo
Jens - JonasDB
Jens - Bellon
Jens - Sander
Dane - Keppens
Dane - Roel
Dane - John
Dane - Onslo
Dane - JonasDB
Dane - Bellon
Dane - Sander
Keppens - Roel
Keppens - John
Keppens - Onslo
Keppens - JonasDB
Keppens - Bellon
Keppens - Sander
Roel - John
Roel - Onslo
Roel - JonasDB
Roel - Bellon
Roel - Sander
John - Onslo
John - JonasDB
John - Bellon
John - Sander
Onslo - JonasDB
Onslo - Bellon
Onslo - Sander
JonasDB - Bellon
JonasDB - Sander
Bellon - Sander

我使用了 Multimap，因为我需要多个具有相同值的键。

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1 回答 1

2

一种比较简单的方法是循环距离，因此我们首先输出距离为 1、2、3 等的所有匹配。

这个的基本版本：

for(int dist = 1; dist < players.length; dist++)
for(int i = 0; i + dist < players.length; i++)
    System.out.println(players[i] + " - " + players[i+dist]);

这将按以下顺序进行匹配：（为简洁起见，按距离分组）

0 - 1, 1 - 2, 2 - 3, 3 - 4, 4 - 5, 5 - 6, 
0 - 2, 1 - 3, 2 - 4, 3 - 5, 4 - 6, 
0 - 3, 1 - 4, 2 - 5, 3 - 6, 
0 - 4, 1 - 5, 2 - 6, 
0 - 5, 1 - 6, 
0 - 6,

如果您想避免第一行中每个人连续玩 2 场比赛的情况，您可以将其分开并按奇数和偶数分开：

for(int i = 0; i < players.length-1; i+=2)
    System.out.println(players[i] + " - " + players[i+1]);
for(int i = 1; i < players.length-1; i+=2)
    System.out.println(players[i] + " - " + players[i+1]);

for(int dist = 2; dist < players.length; dist++)
for(int i = 0; i + dist < players.length; i++)
    System.out.println(players[i] + " - " + players[i+dist]);

这将按以下顺序进行比赛：

0 - 1, 2 - 3, 4 - 5, 
1 - 2, 3 - 4, 5 - 6, 
0 - 2, 1 - 3, 2 - 4, 3 - 5, 4 - 6, 
0 - 3, 1 - 4, 2 - 5, 3 - 6, 
0 - 4, 1 - 5, 2 - 6, 
0 - 5, 1 - 6, 
0 - 6,

对此的一种变体将是环绕并且仅循环超过一半的距离（有一种特殊情况，以避免distance = length/2线重复匹配偶数大小的数组）。

for(int i = 0; i < players.length; i+=2)
    System.out.println(players[i] + " - " + players[(i+1)%players.length]);
for(int i = 1; i < players.length; i+=2)
    System.out.println(players[i] + " - " + players[(i+1)%players.length]);

for(int dist = 2; dist < (players.length+1)/2; dist++)
for(int i = 0; i < players.length; i++)
    System.out.println(players[i] + " - " + players[(i+dist)%players.length]);

if (players.length % 2 == 0)
    for(int i = 0; i < players.length/2; i++)
        System.out.println(players[i] + " - " + players[i+players.length/2]);

比赛将如下所示：

0 - 1, 2 - 3, 4 - 5, 6 - 0, 
1 - 2, 3 - 4, 5 - 6, 
0 - 2, 1 - 3, 2 - 4, 3 - 5, 4 - 6, 5 - 0, 6 - 1, 
0 - 3, 1 - 4, 2 - 5, 3 - 6, 4 - 0, 5 - 1, 6 - 2,
于 2017-04-06T12:28:06.813 回答