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我正在努力学习 Ruby Koans 以尝试学习 Ruby,到目前为止一切顺利。我已经得到了贪婪的公案,在撰写本文时它是 183。我有一个可行的解决方案,但我觉得我只是拼凑了一堆 if/then 逻辑,但我不是拥抱 Ruby 模式。
在下面的代码中,有什么方法可以让我更全面地接受 Ruby 模式吗? (我的代码包含在“我的代码 [BEGINS|ENDS] HERE”注释中。
# Greed is a dice game where you roll up to five dice to accumulate
# points. The following "score" function will be used calculate the
# score of a single roll of the dice.
#
# 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 fives is 500 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:
#
# Your goal is to write the score method.
# MY CODE BEGINS HERE
def score(dice)
# set up basic vars to handle total points and count of each number
total = 0
count = [0, 0, 0, 0, 0, 0]
# for each die, make sure we've counted how many occurrencess there are
dice.each do |die|
count[ die - 1 ] += 1
end
# iterate over each, and handle points for singles and triples
count.each_with_index do |count, index|
if count == 3
total = doTriples( index + 1, total )
elsif count < 3
total = doSingles( index + 1, count, total )
elsif count > 3
total = doTriples( index + 1, total )
total = doSingles( index + 1, count % 3, total )
end
end
# return the new point total
total
end
def doTriples( number, total )
if number == 1
total += 1000
else
total += ( number ) * 100
end
total
end
def doSingles( number, count, total )
if number == 1
total += ( 100 * count )
elsif number == 5
total += ( 50 * count )
end
total
end
# MY CODE ENDS HERE
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_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
end
非常感谢您在我努力了解 Ruby 时提供的任何帮助。
最佳答案
哇!这里有很多非常酷的方法。我喜欢每个人的创造力。但是,我对这里提供的所有答案都有一个教学问题。 (“教育学是对……教学过程的研究。”——维基百科)
从前几个 koans(回到 about_asserts.rb)可以明显看出,启蒙之路不需要任何 Ruby 的先前/外部知识。 Path 甚至不需要先前的编程经验,这一点似乎也很清楚。因此,从教育的角度来看,这个公案必须仅使用早期公案中教授的方法、语言结构和编程技术就可以回答。这意味着:
Enumerable#each_with_index
Enumerable#count
Enumerable#sort
Hash.new(0)
指定默认值Numeric#abs
Numeric#divmod
案例
当
现在,我并不是说您不允许在您的实现中使用这些东西,但公案不能要求使用它们。 必须有一个解决方案,该解决方案仅使用先前 koans 引入的构造。
此外,由于模板只是
def score(dice)
# You need to write this method
end
这似乎暗示解决方案不应定义其他方法或类。也就是说,您应该只替换 # You need to write this method
行。
这是一个符合我的哲学要求的解决方案:
def score (dice)
sum = 0
(1..6).each do |i|
idice = dice.select { |d| d == i }
count = idice.size
if count >= 3
sum += (i==1 ? 1000 : i*100)
end
sum += (count % 3) * 100 if i == 1
sum += (count % 3) * 50 if i == 5
end
sum
end
此处的方法/构造在以下 koan 文件中介绍:
Enumerable#each about_iteration.rb
Enumerable#select about_iteration.rb
Array#size about_arrays.rb
a ? b : c about_control_statements.rb
% about_control_statements.rb
相关 StackOverflow 问题:
关于Ruby Greed Koan - 如何改进我的 if/then 汤?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/4749973/
我正在努力学习 Ruby Koans 以尝试学习 Ruby,到目前为止一切顺利。我已经得到了贪婪的公案,在撰写本文时它是 183。我有一个可行的解决方案,但我觉得我只是拼凑了一堆 if/then 逻辑
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