Question

我有这个python函数代码，假设在任何参数供应中平衡“括号”的数量。虽然，它适用于练习中的需要问题但是当参数包含4个或更多括号时，该功能无法平衡它。下面是代码。

def closedBracket(string_input):
   import  unicodedata, re
   left_bracket = re.compile('[\u0028]')
   right_bracket = re.compile('[\u0029]')
   left_bracket = len(left_bracket.findall(string_input))
   right_bracket = len(right_bracket.findall(string_input))
   if left_bracket > right_bracket:
      # Remove  the leftmost bracket or add one last right bracket
      count = 0
      largest_bracket = (left_bracket - right_bracket)
      while count <= largest_bracket:
         final_string = str(string_input) + '\u0029'
         count += 1
      print('False', final_string)
   elif left_bracket < right_bracket:
      # Remove the last Rightmost Bracket or add another leftmost bracket
      count = 0
      largest_bracket = (right_bracket - left_bracket)
      while count <= largest_bracket:
         final_string = '\u0028' + str(string_input)
         count += 1
      print('False', final_string)

   else:
     print('True')

closedBracket("(5+3)*2)")

The output is "False ((5+3)*2)"

上面的功能对供应参数起作用，但是当我在下面的供应参数上运行时，输出是错误的，所以我需要建议做什么。

def closedBracket(string_input):
   import  unicodedata, re
   left_bracket = re.compile('[\u0028]')
   right_bracket = re.compile('[\u0029]')
   left_bracket = len(left_bracket.findall(string_input))
   right_bracket = len(right_bracket.findall(string_input))
   if left_bracket > right_bracket:
      # Remove  the leftmost bracket or add one last right bracket
      count = 0
      largest_bracket = (left_bracket - right_bracket)
      while count <= largest_bracket:
         final_string = str(string_input) + '\u0029'
         count += 1
      print('False', final_string)
   elif left_bracket < right_bracket:
      # Remove the last Rightmost Bracket or add another leftmost bracket
      count = 0
      largest_bracket = (right_bracket - left_bracket)
      while count <= largest_bracket:
         final_string = '\u0028' + str(string_input)
         count += 1
      print('False', final_string)

   else:
     print('True')

closedBracket("((((5+3)*2)")

The output is "False ((((5+3)*2))" and this is wrong

Answer 1

在循环中你有这一行

final_string = str(string_input) + '\u0029'

这意味着每次你输入输入字符串并添加)。因此，无论您执行多少次迭代，final_string始终等于input_string加入)。

Answer 2

感谢所有来救我的人，特别是@Alik，上述python代码的解决方案在“while循环”部分下解决。 1.此函数名称为closedBracket（），它采用数学表达式，迭代表达式，确定开括号和右括号的数量，如果括号内有不平衡数，则函数将返回数学表达式的平衡版本 2.该功能采用参数

返工完整代码如下：

def closedBracket(string_input):
    import  unicodedata, re
    left_bracket = re.compile('[\u0028]')   # compiling unicode of left
    right_bracket = re.compile('[\u0029]')  # compiling unicode of right
    string_holder = []      # Declaring empty List to hold split string
    string_holder[:0] = string_input # Change the input string to list
    left_bracket = string_holder.count('\u0028')
    right_bracket = string_holder.count('\u0029')
    final_string = []   # Accumulator of
    if left_bracket > right_bracket:
        # Remove  some  leftmost brackets or add more right brackets
        count = 0
        while count < (left_bracket - right_bracket):
            final_string.append('\u0029',) 
            count += 1
        final_string = ''.join(final_string)

        print('False', string_input + final_string)

    elif left_bracket < right_bracket:
        #Remove the last Rightmost Brackets or add another leftmost )
        count = 0
        number_right_bracket = (right_bracket - left_bracket)
        while count < number_right_bracket:
            final_string.append('\u0028',)
            count += 1
        final_string = ''.join(final_string)
        print('False', final_string + string_input)

    else:
        return True

closedBracket("5+((3)*2)))))")

我的代码无法迭代代码的某些部分

2 个答案: