所以我发现我可以通过哈希进行计数,问题是对于7和9,我有四个值。我尝试过其他几件事,但都没有成功。有人能帮助我理解我还能做些什么来从散列中获得我想要的值吗。我意识到我可以将数字与键进行匹配,但我不知道如何使值进行排列。
letters = {"1" => ["1", "1", "1"],
"2" => ["a", "b", "c"],
"3" => ["d", "e", "f"],
"4" => ["g", "h", "i"],
"5" => ["j", "k", "l"],
"6" => ["m", "n", "o"],
"7" => ["p", "q", "r", "s"],
"8" => ["t", "u", "v"],
"9" => ["w", "x", "y", "z"]}
phone_number = gets.chomp.to_s
words = []
word = []
numbers = phone_number.chomp.chars
count0 = 0
while count0 < 3
count1 = 0
while count1 < 3
count2 = 0
while count2 < 3
count3 = 0
while count3 < 3
count4 = 0
while count4 < 3
count5 = 0
while count5 < 3
count6 = 0
while count6 < 3
word[0] = letters[numbers[0]][count0]
word[1] = letters[numbers[1]][count1]
word[2] = letters[numbers[2]][count2]
word[3] = letters[numbers[3]][count3]
word[4] = letters[numbers[4]][count4]
word[5] = letters[numbers[5]][count5]
word[6] = letters[numbers[6]][count6]
words << word.join
count6 += 1
end
count5 += 1
end
count4 += 1
end
count3 += 1
end
count2 += 1
end
count1 += 1
end
count0 += 1
end
puts words
编辑:
我想要一个七位数的数字,并打印出所有可能的字母组合。我是一个初学者,所以我想用我现在知道的东西来理解。我想尝试用if
语句来实现这一点。
numbers = phone_number.chomp.chars
if letters.key?(numbers[0])
if letters.key?(numbers[1])
if letters.key?(numbers[2])
if letters.key?(numbers[3])
if letters.key?(numbers[4])
if letters.key?(numbers[5])
if letters.key?(numbers[6])
end
end
end
end
end
end
end
我知道如何从匹配的键中获取值,但不知道如何在处理其余值时保持第一个值,如果这有意义的话。
product
是您要查找的函数,以下函数适用于任何数字:
digits = '27'
keys = digits.chars.map{|digit|letters[digit]}
p keys.shift.product(*keys).map(&:join) #=> ["ap", "aq", "ar", "as", "bp", "bq", "br", "bs", "cp", "cq", "cr", "cs"]
这将打印可变大小电话号码的所有可能单词:
letters = {"1" => ["1"],
"2" => ["a", "b", "c"],
"3" => ["d", "e", "f"],
"4" => ["g", "h", "i"],
"5" => ["j", "k", "l"],
"6" => ["m", "n", "o"],
"7" => ["p", "q", "r", "s"],
"8" => ["t", "u", "v"],
"9" => ["w", "x", "y", "z"]}
digits = gets.chomp.split ''
# Total number of combinations
n = digits.inject(1) { |a,b| a * letters[b].size }
words = []
0.upto n-1 do |q|
word = []
digits.reverse.each do |digit|
q, r = q.divmod letters[digit].size
word.unshift letters[digit][r]
end
words << word.join
end
puts words
例如,如果输入是67
,则有12种组合:
mp mq mr ms np nq nr ns操作oq或os
编辑:我看不出有什么方法可以像你写的那样使用7个if
语句,但也许这更接近你想要的答案:
words = []
letters[digits[0]].each do |c0|
letters[digits[1]].each do |c1|
letters[digits[2]].each do |c2|
letters[digits[3]].each do |c3|
letters[digits[4]].each do |c4|
letters[digits[5]].each do |c5|
letters[digits[6]].each do |c6|
words << [c0,c1,c2,c3,c4,c5,c6].join
end
end
end
end
end
end
end
puts words
一个好的练习是用一种可以适用于任何长度的电话号码的方式重写它,而不仅仅是7。同样,这只是为了教学目的。在实践中,人们会使用Array的product
方法,就像hirolau的回答一样。
LETTERS = {"1" => ["1", "1", "1"],
"2" => ["a", "b", "c"],
"3" => ["d", "e", "f"],
"4" => ["g", "h", "i"],
"5" => ["j", "k", "l"],
"6" => ["m", "n", "o"],
"7" => ["p", "q", "r", "s"],
"8" => ["t", "u", "v"],
"9" => ["w", "x", "y", "z"]}
def convert_to_phone_number(string)
string.each_char.with_object([]) { |x, arr| LETTERS.each { |k,v| (arr.push k; break) if v.include?(x) }}.join
end
convert_to_phone_number "foobar"
#=> "366227"
我认为这是缓存的问题你需要像下面的一样改变
LETTERS = {"1" => ["1", "1", "1"],
"2" => ["a", "b", "c"],
"3" => ["d", "e", "f"],
"4" => ["g", "h", "i"],
"5" => ["j", "k", "l"],
"6" => ["m", "n", "o"],
"7" => ["p", "q", "r", "s"],
"8" => ["t", "u", "v"],
"9" => ["w", "x", "y", "z"]}
def convert_to_phone_number(string)
string.each_char.with_object([]) { |x, arr| LETTERS.each { |k,v| (arr.push k; break) if v.include?(x) }}.join
end
convert_to_phone_number"foobar"