Algorithm 将字符串从一种格式转换为另一种格式的算法

我正在研究一个问题，该问题声明如下转换字符串

s = "3[a]2[bc]", return "aaabcbc".
s = "3[a2[c]]", return "accaccacc".
s = "2[abc]3[cd]ef", return "abcabccdcdcdef".

我能够理解如何做到这一点

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我在想有没有一种相反的方法。当给定一个像abcabcdcdef这样的字符串时，我知道可以有很多表示的可能性。我在想，我们能不能用内存最少的表示法来表示，不是算法，而是最后一个字符串。

为了获得最大的效率，我们希望尽可能地减少内存。我想我会这样做它可能不是最有效的算法：

s = "whateverwhateveryouwantwantwantababababababababc"
possibilities = []
repeats = []
def findRepeats(repeats, s, length):
    for i in range(0, len(s) - 2 * length + 1):
        if s[i:i+length] == s[i+length:i+2*length]:
            trackInd = i+length
            times = 2
            while trackInd+2*length <= len(s):
                if (s[trackInd:trackInd+length]==s[trackInd+length:trackInd+2*length]):
                    times += 1
                else: break
                trackInd += length

            repeats.append((i, times, s[i:i+length]))

    return repeats

for i in range(0, len(s)):
    repeats = findRepeats(repeats, s, i)

def formPossibility(repeats, s):
    build = ""
    i = 0
    while i < len(s):
        pass = True
        for repeat in repeats:
            if repeat[0] == i:
                pass = False
                build += repeat[1] + "["
                build += repeat[2] + "]"
                break

        if pass:
            build += s[i]

# I didn't finish this but you would loop through all the repeats and test
# them to see if they overlap, and then you would take all the posibilities
# of different ways to make them so that some are there, and some are not.
# in any case, I think that you get the idea.
# I couldn't finish this because I am doing the coding on stackoverflow and
# its like so painful and so hard to debug. also I don't have enough time sorry

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我不知道它是最有效的还是最有效的，但下面是我使用js的方法

函数格式模式、长度、时间{ var结果=； 如果时间==0{ 结果=模式； }否则{ 结果=次数+1.toString+[+模式+]； } 返回结果； } 函数编码输入{ var结果=； var模式={length:1，times:0}； var i=1； 而我是pattern.length*pattern.times{ pattern.length=i； pattern.times=j； } i++； } 如果pattern.length>1{ 结果=formatencodeinput.substr0，pattern.length，pattern.length，pattern.times； }否则{ 结果=formatinput.substr0，pattern.length，pattern.length，pattern.times； } 如果pattern.length+pattern.length*pattern.times}不，不是家庭作业。我正在做。可能重复的检查旧的问题。对于嵌套和方括号，您的情况似乎更为复杂，但AFAICT这种方法也应该适用于您。我在想，这个答案是否适用于嵌套？@user168983它可能可以调整，但代价是变成立方或四次方。