写“匹配平衡括号"的更好方法Ruby 中的程序 [英] Better way to write "matching balanced parenthesis" program in Ruby
问题描述
This method is supposed to take a string and detect if the brackets '(' '{' '[' in the string are closing properly with the corresponding (opposite) brackets.
First, is there a more elegant, compact way to write this bit without using all the "or"s (||):
split_array.each do |i|
if (i == "{" || i == "(" || i == "[")
left.push(i)
else (i == "}" || i == ")" || i == "]")
right.push(i)
end
end
My second question is, is this code terrible (see below)? It seems I should be able to write this in way fewer lines, but logically, I haven't come up with another solution (yet.) The code works for most tests, but it returns false for this test (see all driver tests at bottom): p valid_string?("[ ( text ) {} ]") == true
Any critique would be greatly appreciated! (also, if there is a better section to post this, please let me know) Thanks!
def valid_string?(string)
opposites = { "[" => "]", "{" => "}", "(" => ")", "]" => "[", "}" => "{", ")" => "(" }
left = Array.new
right = Array.new
return_val = true
split_array = string.split(//)
split_array.delete_if { |e| e.match(/s/) }
split_array.each do |i|
if (i == "{" || i == "(" || i == "[")
left.push(i)
else (i == "}" || i == ")" || i == "]")
right.push(i)
end
end
# p left
# p right
left.each_index do |i|
if left[i] != opposites[right[i]]
return_val = false
end
end
return_val
end
p valid_string?("[ ] } ]") == false
p valid_string?("[ ]") == true
p valid_string?("[ ") == false
p valid_string?("[ ( text ) {} ]") == true
p valid_string?("[ ( text { ) } ]") == false
p valid_string?("[ (] {}") == false
p valid_string?("[ ( ) ") == false
-------Updated: After trying some different approaches, my refactor is this:-----------
def valid_string?(str)
mirrored = { "[" => "]", "{" => "}", "(" => ")" }
open_brackets = Array.new
split_str_array = str.split("")
split_str_array.each do |bracket|
if bracket.match(/[[|{|(]/) then open_brackets.push(bracket)
elsif bracket.match(/[]|}|)]/)
return false if mirrored[open_brackets.pop] != bracket
end
end
open_brackets.empty?
end
My approach is as below :
def valid_string?(string)
open_paren = ['[','{','(']
close_paren = [']','}',')']
open_close_hash = {"]"=>"[", "}"=>"{", ")"=>"("}
stack = []
regex = Regexp.union(close_paren+open_paren)
string.scan(regex).each do |char|
if open_paren.include? char
stack.push(char)
elsif close_paren.include? char
pop_val = stack.pop
return false if pop_val != open_close_hash[char]
end
end
open_paren.none? { |paren| stack.include? paren }
end
valid_string?("[ ] } ]") # => false
valid_string?("[ ]") # => true
valid_string?("[ ") # => false
valid_string?("[ (] {}") # => false
valid_string?("[ ( ) ") # => false
valid_string?("[ ( text { ) } ]") # => false
valid_string?("[ ( text ) {} ]") # => true
- Declare a character stack
S
. - Now traverse the expression string exp.
- If the current character is a starting bracket (
‘(‘
or‘{‘
or‘['
) then push it to stack. - If the current character is a closing bracket (
')'
or'}'
or']'
) then pop from stack and if the popped character is the matching starting bracket then fine else parenthesis are not balanced.
- If the current character is a starting bracket (
- After complete traversal, if there is some starting bracket left in stack then "not balanced"
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