python - 给定一个字符串，波兰表示法中最长的 WFF 之一的长度是多少？

Question

我正在尝试 to write Python 中 WFF 'N Proof 游戏(无版权侵权意图)中一直流行的 Count-A-WFF 部分的一个版本。好吧，不太受欢迎。

我想我已经根据需要启动并运行了所有内容，最多可达 4 个字母的字符串。

def maximum_string(s):
if cs(s) == True:
    return len(s)
elif len(s) == 2:
    l1 = [cs(s[0]), cs(s[1])]
    if True in l1:
        return len(s) - 1
    else:
        return 0
elif len(s) == 3:
    first = s[0] + s[1]
    second = s[0] + s[2]
    third = s[1] + s[2]
    l1 = [cs(first), cs(second), cs(third)]
    if True in l1:
        return len(s) - 1
    l2 = [cs(s[0]), cs(s[1]), cs(s[2])]
    if True in l2:
        return len(s) - 2
    else:
        return 0
elif len(s) == 4:
    first = s[0]+s[1]+s[2]
    second = s[0]+s[1]+s[3]
    third = s[1]+s[2]+s[3]
    fourth = s[0]+s[2]+s[3]
    l1 = [cs(first), cs(second), cs(third), cs(fourth)]
    if True in l1:
        return 3
    first = s[0] + s[1]
    second = s[0] + s[2]
    third = s[0] + s[3]
    fourth = s[1] + s[2]
    fifth = s[1] + s[3]
    sixth = s[2] + s[3]
    l2 = [cs(first), cs(second), cs(third), cs(fourth), cs(fifth), cs(sixth)]
    if True in l2:
        return 2
    first = s[0]
    second = s[1]
    third = s[2]
    fourth = s[3]
    l3 = [cs(first), cs(second), cs(third), cs(fourth)]
    if True in l3:
        return 1
    else:
        return 0

def cs(string):
global length_counter, counter, letter
counter = 1
length_counter = 0
letters_left = len(string)
while letters_left != 0 and length_counter < len(string):
    letter = string[length_counter]
    if letter == 'C' or letter == 'A' or letter == 'K' or letter == 'E' or letter == "K":
        counter += 1 
    elif letter == 'N':
        counter += 0
    else:
        counter -= 1  
    length_counter += 1
    letters_left -= 1
if counter == 0 and len(string) == length_counter:
    return True
else:
    return False

maximum_string 辅助函数的目的是，给定任何字符串 S，找到最长可能的 wffs 之一的长度，您可以仅从 S 的字母组成。当然，我可以继续我目前的模式最大长度为 13 的 maximum_string 辅助函数。但是，组合爆炸是显而易见的。那么，有没有更优雅的方法来完成最大字符串辅助函数？

Answer 1

实际上，我之前使用的其中一个函数会返回一个字符串与波兰表示法排列相距多远的距离。因此，这比我预期的更容易修复。这就是我要找的东西:

def maximum_string(string):
    global length_counter, counter, letter
    counter = 1
    length_counter = 0
    letters_left = len(string)
    while letters_left != 0 and length_counter < len(string):
        letter = string[length_counter]
        if letter == 'C' or letter == 'A' or letter == 'K' or letter == 'E' or letter == "K":
            counter += 1 
        elif letter == 'N':
            counter += 0
        else:
            counter -= 1  
        length_counter += 1
        letters_left -= 1
    if ('p' in string) or ('q' in string) or ('r' in string) or ('s' in string) or ('t' in string) or ('u' in string):
        return len(string) - abs(counter)
    else:
        return 0