Python程序检查简单括号的匹配 [英] Python program to check matching of simple parentheses

问题描述

我是Python的新手，我遇到了一个检查给定字符串中的简单括号（，）是否均匀匹配的练习。

<我在这里看到了使用尚未遇到的stack命令的示例，因此我尝试了另一种方法。谁能告诉我我要去哪里哪里？

  def匹配（str）：
 ope = [] 
 clo = [] 
对于范围（0，len（str））中的i：
l = str [i] 
如果l ==（：
 ope = ope + [（] 
 else：
，如果l ==）：
 clo = clo + [）] 
 else：
 return（ ope，clo）如果len（ope）== len（clo），则
：
返回True 
否则：
 return False 
  

想法是将（和）堆积到两个单独的列表中，然后比较列表的长度。我还有另一个版本，我在列表ope和clo后面分别附加了（或）对应的i。

感谢您的时间！

解决方案

下面是一种更为优雅的方法。它清除for循环，并用简单的计数器变量替换列表。如果计数器降到零以下，它也会返回false，这样 matched（）（）将返回 False 。

  def匹配（str）：
 count = 0 
 for str in：
如果i ==（：
计数+ = 1 
 elif i ==）：
计数-= 1 
如果count ＜ 0：
返回False 
返回计数== 0 
  

I am a Python newbie and I came across this exercise of checking whether or not the simple brackets "(", ")" in a given string are matched evenly.

I have seen examples here using the stack command which I havent encountered yet.So I attempted a different approach. Can anyone tell me where I am going wrong?

def matched(str):
    ope = []
    clo = []
    for i in range(0,len(str)):
        l = str[i]
        if l == "(":
            ope = ope + ["("]
        else:
            if l == ")":
                clo = clo  + [")"]
            else:
                return(ope, clo)
    if len(ope)==len(clo):
        return True
    else:
        return False

The idea is to pile up "(" and ")" into two separate lists and then compare the length of the lists. I also had another version where I had appended the lists ope and clo with the relevant i which held either ( or ) respectively.

Thanks for your time!

解决方案

A very slightly more elegant way to do this is below. It cleans up the for loop and replaces the lists with a simple counter variable. It also returns false if the counter drops below zero so that matched(")(") will return False.

def matched(str):
    count = 0
    for i in str:
        if i == "(":
            count += 1
        elif i == ")":
            count -= 1
        if count < 0:
            return False
    return count == 0

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