class Anagram
def self.resolves?(first_word, second_word)
@letter_counts = {}
# Assuming that anagrams aren't case sensitive.
first_word.downcase!
second_word.downcase!
count_letters(first_word, 1)
count_letters(second_word, -1)
(sum_of_counts == 0)
end
private
def self.sum_of_counts
@letter_counts.inject(0){|sum, n| sum + n[1]}
end
def self.count_letters(aword, incrementer)
aword.each_byte {|c| @letter_counts[c].nil? ? @letter_counts[c] = 1 : @letter_counts[c]+=incrementer }
end
end
puts Anagram.resolves?('dog','cat')
puts Anagram.resolves?('dog','god')
puts Anagram.resolves?('Jim','Jones')
puts Anagram.resolves?('Hello','olleH')
Refactorings
No refactoring yet !
darkleo.myopenid.com
February 19, 2010, February 19, 2010 22:01, permalink
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and what wrong with reverse ?
"stressed".reverse => "desserts"
steenslag
February 19, 2010, February 19, 2010 22:45, permalink
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An anagram is not a palindrome, so the .reverse is not of much use.
class String
def anagram?(other)
self.downcase.split(//).sort == other.downcase.split(//).sort
end
end
puts "dog".anagram?("ogd")
bob
February 20, 2010, February 20, 2010 02:43, permalink
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just sort the chars
class Anagram
def self.resolves?(first_word, second_word)
first_word.split('').sort() == second_word.split('').sort()
end
end
puts Anagram.resolves?('dog','cat')
puts Anagram.resolves?('dog','god')
puts Anagram.resolves?('Jim','Jones')
puts Anagram.resolves?('Hello','olleH')
sdotsen
February 23, 2010, February 23, 2010 14:39, permalink
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steenslag is correct, what you're looking for is his code. What you are asking for is a palindrome NOT an anagram.
This code outputs whether two strings resolve to being an anagram or not.
I particularly don't like the count_letters method and the checking of the existence of the hash value. Is there a better way?