Question

我有一个团队阵列：

var teams = [a, b, c, d]

我需要一个函数来生成所有可能的匹配对，例如，如果a与b一起播放，则c会以d格式播放此格式：

[{ team1: a, team2: b }, { team1: c, team2: d }]

然后4支球队的最终结果如下：

var matches = {
    0: [{ team1: a, team2: b }, { team1: c, team2: d }]
    1: [{ team1: b, team2: c }, { team1: a, team2: d }]
    2: [{ team1: c, team2: a }, { team1: d, team2: b }]
}

团队数量始终为偶数且大于3（4,6,8，...）

Answer 1

一种简单的方法，迭代直到每支球队都与他的所有敌人作战：

 var teams = ["a", "b", "c", "d"];
teams = teams.map(id => ({id}));
teams.forEach(team => team.enemies = teams.filter(enemy => enemy !== team));

const matches = [];

while(teams.some(team => team.enemies.length)){
 const playing = [];
 for(const team of teams){
  if(playing.includes(team)) continue;
   const enemy = team.enemies.find(enemy => !playing.includes(enemy));
   if(!enemy) continue;
   team.enemies.splice(team.enemies.indexOf(enemy),1);
   enemy.enemies.splice(enemy.enemies.indexOf(team), 1);
   //console.log(team.id, enemy.id, playing.map(t => t.id));
   playing.push(team, enemy);
 }
 if(playing.length) matches.push(playing.map(t => t.id))
}

console.log(matches);

每场比赛可以将比赛分成两队

Answer 2

我认为这是一个非重复对的一般问题。

var teams = ["a", "b", "c", "d"]

function possibleMatches(teams) {
  for (var firstTeam = 0; firstTeam < teams.length; firstTeam++) {
    for (var secondTeam = firstTeam + 1; secondTeam < teams.length; secondTeam++) {
      console.log("Your firstTeam and second Team", teams[firstTeam], teams[secondTeam])
    }
  }

}

possibleMatches(teams)

足球夹具javascript算法

2 个答案: