红宝石 - 岩石,纸.剪刀比赛

Question

我需要实现一个摇滚,纸张,剪刀锦标赛模拟器,它将播放所有轮次并返回锦标赛.

这是我的锦标赛阵列:

tournament = [
    [
        [ 
            ["Armando", "P"], ["Dave", "S"] 
        ],
        [ 
            ["Richard", "R"],  ["Michael", "S"] 

        ],
    ],
    [
        [ ["Allen", "S"], ["Omer", "P"] ],
        [ ["David E.", "R"], ["Richard X.", "P"] ]
    ]
]

例如["Richard","R"],["Michael","S"]代表理查德饰演Rock和Michael Scissors的游戏.理查德应该是胜利者并进入下一轮.

我正在尝试为此实现递归函数,但我无法正确获取数组

这是我的方法:

 def rps_tournament_winner(tournament)

  qualifying_round_winners = []
   round_winners = []
  # last round

  if tournament.length < 1
    return
  end

  tournament.each_with_index do |round,i|   
 puts round.inspect
     qualifying_round_winners[i] = []
     round_winners = []
     round.each_with_index do |game,j|
         winner = rps_game_winner(game)
         round_winners.push winner
     end
     qualifying_round_winners[i] = round_winners

  end

  rps_tournament_winner(qualifying_round_winners)
end

我已经为排位赛的胜利者发明了很多阵容但是无法找到解决方案.希望您能够帮助我.谢谢.

编辑:

这是我在方法的每次迭代中应该得到的

在第一轮之后

tournament = [
    [
        [ 
            ["Dave", "S"], ["Richard", "R"] 
        ],
        [ 
            ["Allen", "S"],  ["Richard X.", "P"] 

        ]
    ]
]

第二轮后:

tournament = [
    [
        [ 
            ["Richard", "R"] , ["Allen", "S"]
        ]
    ]
]

优胜者

["Richard", "R"] 

编辑:2

这是完整的代码:http: //pastebin.com/gjKfiWLD

Answer 1

在我们推算时,我们必须考虑两种情况:

rps_tournament_winner的论点是游戏,或者rps_tournament_winner的论点是锦标赛.请注意,锦标赛的子部分本身就是一个锦标赛(仅少一轮).如果这是一场比赛我们会赢回冠军.如果这是一个小节/锦标赛,我们会努力寻找其两个孩子的获胜者.

def rps_tournament_winner(tournament)
    # Check if we're at a game
    if tournament[0][0].is_a? String
        return rps_game_winner(tournament)
    end
    # Otherwise keep going down the rabbit hole
    return rps_game_winner([rps_tournament_winner(tournament[0]),rps_tournament_winner(tournament[1])])
end

这有助于您理解吗？