查找外部括号的计数 [英] Finding the count of external Parenthesis

问题描述

我正在试图找出找到外括号计数的解决方案。

您将在下面的示例中更清楚地了解外部括号的含义。假设给定的表达式始终有效。

I am trying to figure out the solution for finding the count of external parenthesis.
You will have more clarity on what i mean by external parenthesis in the below examples. Assume that the given expression will always be valid.

(a+b)                   //External parenthesis count =  1
(a+b)*(a-b)             //External parenthesis count =  0 
((a+b)*(a-b))           //External parenthesis count =  1
(((a+b)*(a-b)))         //External parenthesis count =  2
((a))                   //External parenthesis count =  2
(a + (b*c))             //External parenthesis count =  1
(a+b)+c                 //External parenthesis count =  0

我试过这个并且不知何故无法得到正确的逻辑。

所以我正在寻求创意。我从逻辑的角度来看更多。

I have tried this and somehow not able to get the right logic.
So i am reaching out for ideas. I am looking more from a logic perspective.

推荐答案

(a+b)                   //External parenthesis count =  1
(a+b)*(a-b)             //External parenthesis count =  0 
((a+b)*(a-b))           //External parenthesis count =  1
(((a+b)*(a-b)))         //External parenthesis count =  2
((a))                   //External parenthesis count =  2
(a + (b*c))             //External parenthesis count =  1
(a+b)+c                 //External parenthesis count =  0

对于每个（）对删除（包括括号）不包含任何括号本身，例如（a + b）不包含（a + （ b * c ））。

For each ( ) pair delete ones (including the parenthesis) that don't contain any parenthesis themselves, eg (a+b) doesn't but (a + (b*c)) does.

                   //External parenthesis count =  1
*             //External parenthesis count =  0 
(*)           //External parenthesis count =  1
((*))         //External parenthesis count =  2
()                   //External parenthesis count =  2
(a + )             //External parenthesis count =  1
+c                 //External parenthesis count =  0

现在只计算左边的括号对。请注意，这不适用于（a + b）或（（a）），因此当括号仅存在于开头和结尾时，您可能需要一个特殊情况。

Now just count the parenthesis pairs left. Note this doesn't work for (a+b) or ((a)) so you might need a special case for when parenthesis only exists at the beginning and end.

externalCount = 0
loop
  if first or last chars are not parens
    exit loop
  Remove the external pair of parentheses.
  If the result is not well-formed
    exit loop
  externalCount++
  repeat loop
end loop

确定它是否格式正确：

To determine if it is well-formed:

counter = 0
scan across the expression:
  if '(' then counter++
  if ')' then counter--
well formed if at end of the scan the counter is 0 and never went negative.

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