将字符串从一种格式转换为另一种格式的算法答案

【问题标题】：Algorithm to convert string from one format to other将字符串从一种格式转换为另一种格式的算法
【发布时间】：2017-02-02 13:09:47
【问题描述】：

我正在查看一个问题，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。

我在想有没有办法在 reverse 中做到这一点。当给定一个像abcabccdcdcdef 这样的字符串时，我知道可以有很多表示的可能性。我在寻找我们可以用占用最低内存的表示来做到这一点（不是算法，而是最终字符串）。

【问题讨论】：

不，不是作业。我在做this。
Finding the minimum length RLE的可能重复
查看较早的问题。您的情况似乎更复杂（带有嵌套和括号），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 它可能可以适应，但代价是变成三次或四次。

标签： algorithm

【解决方案1】：

为了最大限度地提高效率，我们希望尽可能减少。我想我会做这样的事情（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

【讨论】：

【解决方案2】：

不知道是最高效还是根本不高效，但这是我使用js的方法。

function format(pattern, length, times) {
    var result = "";
    if (times == 0) {
        result = pattern;
    } else {
        result = (times + 1).toString() + "[" + pattern + "]";
    }
    return result;
}

function encode(input) {
    var result = "";
    var pattern = { length: 1, times: 0 };
    var i = 1;
    while (i <= input.length / 2) {
        var subpattern = input.substr(0, i);
        var j = 0;
        while (input.substr(i + j * i, i) == subpattern && j + i < input.length) {
            j++;
        }
        if (i * j > pattern.length * pattern.times) {
            pattern.length = i;
            pattern.times = j;
        }
        i++;
    }
    if (pattern.length > 1) {
        result = format(encode(input.substr(0, pattern.length)), pattern.length, pattern.times);
    } else {
        result = format(input.substr(0, pattern.length), pattern.length, pattern.times);
    }
    if (pattern.length + pattern.length * pattern.times < input.length) {
        result += encode(input.substr(pattern.length + pattern.length * pattern.times, input.length));
    }
    return result;
}

【讨论】：