Question

我正在尝试用Ruby编写游戏Mastermind的版本。为此，我需要编写一个方法，该方法将返回两个不同数组共有多少个元素的计数，但这些元素位于不同的位置。我已经有一个方法可以返回在同一索引处正确的项目数。

例如：

comparing ["red", "green", "red", "orange"] to ["blue", "red", "blue", "blue"]
#should return 1

comparing ["red", "red", "orange", "orange"] to ["orange", "orange", "red", "red"]
#should return 4

comparing ["green", "green", "orange", "blue"] to ["green", "green", "orange", "red"]
#should return 0

Answer 1

def f a, b
  c = [a, b].transpose.inject [] do |m, e|
    m << e if e.first != e.last
    m
  end.transpose
  puts c.first.size - (c.last - c.first).size
end

f [:red, :green, :red, :orange],    [:blue, :red, :blue, :blue]     # => 1
f [:red, :red, :orange, :orange],   [:orange, :orange, :red, :red]  # => 4
f [:green, :green, :orange, :blue], [:green, :green, :orange, :red] # => 0

# it works with Strings, also

f %w/red green red orange/,    %w/blue red blue blue/     # => 1
f %w/red red orange orange/,   %w/orange orange red red/  # => 4
f %w/green green orange blue/, %w/green green orange red/ # => 0

Answer 2

我的方式与@DigitalRoss不同。

def match_at_wrong_position(master, guess)
  guess = guess.dup                          # dup, so we do not modify the input param
  master.each_with_index.inject(0) do |count, (item, index)|
    found_at = guess.index(item)             # is it provided as a guess
    next count if found_at.nil?              # not found
    guess[found_at] = nil                    # make sure it is not found again
    found_at != index ? count + 1 : count    # if not same position, then increment
  end
end

p match_at_wrong_position(["red", "green", "red", "orange"], ["blue", "red", "blue", "blue"])
p match_at_wrong_position(["red", "red", "orange", "orange"], ["orange", "orange", "red", "red"])
p match_at_wrong_position(["green", "green", "orange", "blue"], ["green", "green", "orange", "red"])

Answer 3

只需要发明一种方法来计算正确的颜色（不要介意位置），然后减去相同指数（你已经拥有）的正确颜色数。

如何编写一个ruby方法，返回两个数组但位于不同位置的项目计数

3 个答案: