我想到了这种类似于Java / C#的递归方法。它通过了所有提供的单元测试,因此从技术上讲是正确的。
现在我想知道:这是好的Ruby代码吗?如果不是,那么“ Rubyist”将如何编写此方法?可能:为什么?对于重复代码的数量,我也不太满意,但我想不出更好的Rubyish方式。
def score(dice) #dice is an array of numbers, i.e. [3,4,5,3,3]
return 0 if(dice == [] || dice == nil)
dice.sort!
return 1000 + score(dice[3..-1]) if(dice[0..2] == [1,1,1])
return 600 + score(dice[3..-1]) if(dice[0..2] == [6,6,6])
return 500 + score(dice[3..-1]) if(dice[0..2] == [5,5,5])
return 400 + score(dice[3..-1]) if(dice[0..2] == [4,4,4])
return 300 + score(dice[3..-1]) if(dice[0..2] == [3,3,3])
return 200 + score(dice[3..-1]) if(dice[0..2] == [2,2,2])
return 100 + score(dice[1..-1]) if(dice[0] == 1)
return 50 + score(dice[1..-1]) if(dice[0] == 5)
return 0 + score(dice[1..-1]);
end
某些背景(如果需要的话) )
# Greed is a dice game where you roll up to five dice to accumulate
# points. A greed roll is scored as follows:
#
# * A set of three ones is 1000 points
#
# * A set of three numbers (other than ones) is worth 100 times the
# number. (e.g. three fours is 400 points).
#
# * A one (that is not part of a set of three) is worth 100 points.
#
# * A five (that is not part of a set of three) is worth 50 points.
#
# * Everything else is worth 0 points.
#
#
# Examples:
#
# score([1,1,1,5,1]) => 1150 points
# score([2,3,4,6,2]) => 0 points
# score([3,4,5,3,3]) => 350 points
# score([1,5,1,2,4]) => 250 points
#
# More scoring examples are given in the tests below:
class AboutScoringProject < EdgeCase::Koan
def test_score_of_an_empty_list_is_zero
assert_equal 0, score([])
end
def test_score_of_a_single_roll_of_5_is_50
assert_equal 50, score([5])
end
def test_score_of_a_single_roll_of_1_is_100
assert_equal 100, score([1])
end
def test_score_of_a_single_roll_of_1_is_100
assert_equal 200, score([1,1])
end
def test_score_of_multiple_1s_and_5s_is_the_sum_of_individual_scores
assert_equal 300, score([1,5,5,1])
end
def test_score_of_single_2s_3s_4s_and_6s_are_zero
assert_equal 0, score([2,3,4,6])
end
def test_score_of_a_triple_1_is_1000
assert_equal 1000, score([1,1,1])
end
def test_score_of_other_triples_is_100x
assert_equal 200, score([2,2,2])
assert_equal 300, score([3,3,3])
assert_equal 400, score([4,4,4])
assert_equal 500, score([5,5,5])
assert_equal 600, score([6,6,6])
end
def test_score_of_mixed_is_sum
assert_equal 250, score([2,5,2,2,3])
assert_equal 550, score([5,5,5,5])
end
def test_score_of_a_triple_1_is_1000A
assert_equal 1150, score([1,1,1,5,1])
end
def test_score_of_a_triple_1_is_1000B
assert_equal 350, score([3,4,5,3,3])
end
def test_score_of_a_triple_1_is_1000C
assert_equal 250, score([1,5,1,2,4])
end
end
#1 楼
该代码存在一些问题:如果未将
== nil
指定为该方法的有效值,则不要检查。在这里,检查并返回0可能会掩盖另一个问题。除非必要,请不要使用return语句。在ruby中,几乎所有内容都是一个表达式,并且方法返回最后一个表达式的值。在这里,您可以使用
if...elsif
或case
来代替一系列if
语句。 br /> 以下是应用上述建议的代码版本:
def score(dice)
score = 0
counts = dice.each_with_object(Hash.new(0)) { |x, h| h[x] += 1 }
(1..6).each do |i|
if counts[i] >= 3
score += (i == 1 ? 1000 : 100 * i)
counts[i] = [counts[i] - 3, 0].max
end
score += counts[i] * (i == 1 ? 100 : 50)
end
score
end
评论
\ $ \ begingroup \ $
递归并非直截了当。我发现OP的方法很简单-只是非常重复。
\ $ \ endgroup \ $
–sepp2k
2011年1月29日19:34
\ $ \ begingroup \ $
通常,这与直截了当相反。但是,在这种情况下,我相信是这样。
\ $ \ endgroup \ $
–glebm
2011年1月29日19:42
\ $ \ begingroup \ $
自从开始使用红宝石以来,实际上我已经放弃了每种方法的“只有一个退货”的规则。我经常在一个方法中有多个返回语句。
\ $ \ endgroup \ $
–巴里·赫斯(Barry Hess)
2011年1月29日20:05
\ $ \ begingroup \ $
我所说的不是多个返回点,而是return关键字的用法
\ $ \ endgroup \ $
–glebm
11年1月29日在21:08
\ $ \ begingroup \ $
很棒的评论。这使我陷入一个错误,在我阅读它后必须修复。 :)
\ $ \ endgroup \ $
– Michelle Tilley
2011-02-25 7:44
#2 楼
@glebm有一些非常好的观点。我还想介绍一种不同的风格。这是解决这个问题的方法。def score dice
dice.group_by(&:to_i).inject(0) do |score, combo|
score + combos_score(*combo) + ones_score(*combo) + fives_score(*combo)
end
end
def combos_score dice_value, dice_with_value
number_of_bonues = [dice_with_value.size - 2, 0].max
bonus_for(dice_value) * number_of_bonues
end
def bonus_for dice_value
dice_value == 1 ? 1000 : dice_value * 100
end
def ones_score dice_value, dice_with_value
return 0 if dice_value != 1 || dice_with_value.size > 2
dice_with_value.size * 100
end
def fives_score dice_value, dice_with_value
return 0 if dice_value != 5 || dice_with_value.size > 2
dice_with_value.size * 50
end
我喜欢
每种评分方案的逻辑都被隔离了
无需构建用于计算分数的特殊哈希。
使用内置的Enumerable#group_by可以将相似的骰子聚集在一起
易于测试的小方法
/>
评论
该代码的第二个修订版对我来说看起来不错。使用散列是一个好主意(尽管它没有利用x等于1的[x,x,x]的分数是x * 100,但我猜有5个数字会更好)噪音大于帮助)。