给定数组的组合，在不同的数组中对它们进行排序，这样就不会有重复项

Question

我在给孩子们做足球锦标赛经理。可以有多轮比赛，每轮比赛中的每一支球队都会相互比赛。给出每一轮的组合，我需要比赛的日历，其中最多的球队同时比赛。有没有一种方法可以对组合数组进行排序以获得结果？

我试着“做”一些算法，但是我找不到最好的。在提供的代码中，可以保证两支球队可以同时比赛，但不能超过...

teams = [1,2,3,4,5,6,7,8] # number of teams is variable

pairings = teams.combination(2).to_a
pairings.shuffle!
calendar = []
calendar << pairings.slice!(0)

while pairings.any?
    p = calendar.slice(-1)
    a = p[0]
    b = p[1]
    matched = false
    pairings.each do |pairing|
        next if pairing.include? a
        next if pairing.include? b
        matched = true
        calendar << pairings.delete(pairing)
        break
    end
    unless matched
        p1 = pairings.slice!(0)
        prev = false
        calendar.each do |pairing|
            if pairing.include?(a) || pairing.include?(b)
                prev = true
                next
            end
            unless prev
                i = calendar.index(pairing)
                calendar.insert(i, p1)
                break
            end
            prev = false
        end
    end
end

Answer 1

在谷歌上搜索了我建议的术语后，我找到了这个解决方案，非常感谢！

teams = [1,2,3,4,5]

teams.push "x" if teams.length.odd?

first = [teams.slice!(0)]

teams.length.times do
    half = Integer(teams.length / 2)
    row1 = first + teams.slice(0...half)
    row2 = teams.slice(half...teams.length).reverse
    for i in 0..half do
        next if row1[i] == "x"
        next if row2[i] == "x"
        puts "#{row1[i]} - #{row2[i]}"
    end
    teams.unshift(teams.pop)
end