方法未在Ruby中返回预期值
我试图用Ruby实现Karatsuba乘法方法未在Ruby中返回预期值,ruby,algorithm,Ruby,Algorithm,我试图用Ruby实现Karatsuba乘法 # takes two integer x and y and partition them to x=a+b and y=c+d # example if x = 1234 a=12 and b=34 # recursively compute a*c,a*d,b*c and b*d def mult (x,y) if len(x) == 1 && len(y) == 1 return x*y
# takes two integer x and y and partition them to x=a+b and y=c+d
# example if x = 1234 a=12 and b=34
# recursively compute a*c,a*d,b*c and b*d
def mult (x,y)
if len(x) == 1 && len(y) == 1
return x*y
elsif len(x) > 1 && len(y) > 1
ab = partition(x)
cd = partition(y)
return ab.product(cd).each{ |num| mult(num[0],num[1]) }
end
end
#method for partitioning works fine..
def partition(number)
number.divmod( 10**(len(number)/2) )
end
#method to find size of integer works fine...
def len(value)
value.to_s.split("").compact.size
end
那么
p mult(12,34) should be 3,4,6,8
but is [[1, 3], [1, 4], [2, 3], [2, 4]]
当我在
行号3
中使用打印“#{x*y}”时,它会打印3,4,6,8,而不是返回x*y
。我无法理解为什么mult
方法为x*y
返回nil
问题在于迭代器错误:
# ⇓⇓⇓⇓
ab.product(cd).each{ |num| mult(num[0],num[1]) }
您想要的是:
旁注:您也不需要显式调用return
:
def mult (x,y)
if len(x) == 1 && len(y) == 1
x*y
elsif len(x) > 1 && len(y) > 1
ab = partition(x)
cd = partition(y)
ab.product(cd).map { |num| mult(num[0], num[1]) }
else
raise "We got a problem"
end
end
#method for partitioning works fine..
def partition(number)
number.divmod( 10**(len(number)/2) )
end
#method to find size of integer works fine...
def len(value)
value.to_s.size
end
p mult 12, 34
#⇒ [3,4,6,8]
问题是迭代器错误:
# ⇓⇓⇓⇓
ab.product(cd).each{ |num| mult(num[0],num[1]) }
您想要的是:
旁注:您也不需要显式调用return
:
def mult (x,y)
if len(x) == 1 && len(y) == 1
x*y
elsif len(x) > 1 && len(y) > 1
ab = partition(x)
cd = partition(y)
ab.product(cd).map { |num| mult(num[0], num[1]) }
else
raise "We got a problem"
end
end
#method for partitioning works fine..
def partition(number)
number.divmod( 10**(len(number)/2) )
end
#method to find size of integer works fine...
def len(value)
value.to_s.size
end
p mult 12, 34
#⇒ [3,4,6,8]
len(x)
?听起来像Python@Ursus这听起来像是自我实现的方法,检查代码片段的最后3行。len(x)
?听起来像Python@Ursus这听起来像是自我实现的方法,检查代码段的最后3行。感谢它的工作,我应该知道它何时打印出[[1,3],[1,4],[2,3],[2,4]
并停止,但是当我必须迭代每个值并将其传递给mult(x,y)时,为什么每个都不工作呢方法。因为每个都返回iterable本身,而映射实际上将iterable转换为从mult
方法返回的新值。感谢它的工作,我应该知道它何时打印[[1,3],[1,4],[2,3],[2,4]]
和stopped但是当我必须迭代每个值并将其传递给mult(x,y)
方法时,每个都会返回iterable本身,而映射实际上会将iterable转换为从mult
方法返回的新值。