长期解决方案

Question

我试图在阵列中一次抓取两个子阵列 - 关于如何做到这一点的任何想法？

示例：

deck = [[2,spades],[3,hearts],[6,diamonds],[10,clubs],[8,hearts],[9,clubs]]

有没有办法一次抓住两张牌，所以索引每次都会增加，然后从[0..1]再到[1..0]，依此类推，抓住两张牌时间。

我想说：

玩家1卡：[2,spades],[3,hearts]

玩家2卡：[6,diamonds],[10,clubs]

Answer 1

我的建议只是将deck数组传递给方法并返回一个带[player1，player2，deck]的数组。如果你只是从牌组的“顶部”绘制，你可以使用shift从阵列中取出第一个元素。

长期解决方案

deck = [[2,"spades"],[3,"hearts"],[6,"diamonds"],[10,"clubs"],[8,"hearts"],[9,"clubs"]]

def drawTwo(arr)
  if arr.count >= 4
    player_one = [deck.shift, deck.shift]
    player_two = [deck.shift, deck.shift]
    return [player_one, player_two, deck]
  else
    return "Not enough cards in deck, please provide a new deck"
  end
end

round = drawTwo(deck)
player_one = round[0]
player_two = round[1]
deck = round[2]

puts "Player one: #{player_one}"
puts "Player two: #{player_two}"
puts "Deck: #{deck}"

我试着非常详细，并没有对这段代码进行过多的混淆，所以它应该读起来非常明确。

你可以通过这样重写它来缩短它，我只是想让它变得可以理解发生的事情：

浓缩溶液

deck = [[2,"spades"],[3,"hearts"],[6,"diamonds"],[10,"clubs"],[8,"hearts"],[9,"clubs"]]

def drawTwo(arr)
  arr.count >= 4 ? [[arr.shift, arr.shift], [arr.shift, arr.shift]] : raise "Not enough cards..."
end

player_one, player_two = drawTwo(deck)

puts "Player one: #{player_one}"
puts "Player two: #{player_two}"
puts "Deck: #{deck}"

首次生成套牌时，请务必添加deck.shuffle。

另外，我不知道你用什么来制作套牌，但是因为我玩得很开心：

生成混洗甲板

def newShuffledDeck
    ranks = ["2", "3", "4", "5", "6", "7", "8", "9", "10", "J", "Q", "K", "A"]
    suits = ["hearts", "spades", "clubs", "diamonds"]
    deck = ranks.product(suits)
    deck.shuffle
end

Answer 2

试试这个

hands = array.each_slice(hand_size).take(player_count)

hands.each_with_index do |hand, n|
  puts "player #{n + 1} cards are: #{hand}"
end

这是如何运作的？

each_slice(n)将数组切割成长度为n
take(n)获取第一个n件

Answer 3

您可以使用Enumerable::each_slice

deck = [[2,"spades"],[3,"hearts"],[6,"diamonds"],[10,"clubs"],[8,"hearts"],[9,"clubs"]]

deck.each_slice(2) do |el|
  puts el
end

你如何以增量迭代2d数组并用ruby增加增量？

3 个答案:

长期解决方案

浓缩溶液

生成混洗甲板