我的主要问题是尝试找到一个合适的解决方案来自动转换它，例如：
d+c+d+f+d+c+d+f+d+c+d+f+d+c+d+f+
进入这个：
[d+c+d+f+]4
即。找到彼此相邻的重复项，然后从这些重复项中缩短“循环”。 到目前为止，我找不到合适的解决方案，我期待着回应。附：为了避免混淆，前面提到的样本并不是唯一需要“循环”的东西，它因文件而异。哦，这是用于C ++或C＃程序，要么很好，尽管我也接受任何其他建议。此外，主要思想是所有工作都将由程序本身完成，除了文件本身之外没有用户输入。 这是完整的文件，供参考，我对拉伸页面表示歉意： ＃0 @ 16 v225 y10 w250 t76
L16 $ ED $ EF $ A9 p20,20 ＆GT; ecegb＆GT; d＆LT; bgbgecgec＆LT;克 ＆GT; d +＆LT b取代; d + F + A +＆以及c +＆LT; A + F + A + F + d + LT b取代; F + d + LT; BF + ＆以及c＆LT a取代; cegbgegec＆LT a取代; EC＆LT; AE ＆GT; d + C + d + F + d + C + d + F + d + C + d + F + d + C + d + F + R1 ^ 1
/ L8 r1r1r1r1 F +＆LT; A +＆GT; F + G + CG + R4 A + C + A + G + CG + R4F + LT; A +＆GT; F + G + CG + R4 A + C + A + G + CG + R4F + LT; A +＆GT; F + G + CG + R4 A + C + A + G + CG + R4 F +＆LT; A +＆GT; F + G + CG + R4 A + C + A + G + r4g + 16f16c + 一个+ 2 ^ G + F + G + 4 F + FF + 4FD + F4 d + C + d + 4C +℃下A + 2 ^ 4 ＆GT; C4D + ＆LT; G + 2 ^ 4R4 ^ 一个+以及c + d + 4G + 4A + 4 R1 ^ 2 ^ 4 ^一个+ 2 ^ G + F + G + 4 F + FF + 4FD + F4 d + C + d + 4C +℃下A + 2 ^ 4 ＆GT; C4D + ＆LT; G + 2 ^ 4R4 ^ 一个+以及c + d + 4G + 4A + 4 R1 ^ 2 ^ 4 ^ r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1
＃4 @ 22 v250 y10
L8 O3 RG + RG + RG + RG + RG + RG + RG + RG + RG + RG + RG + RG + RG + RG + RG + RG + RG + RG + RG + RG + RG + RG + RG + RG + / r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1
＃2 @ 4 v155 y10
L8 $ ED $ F8 $ 8F O4 r1r1r1 d + 4f4f + 4G + 4 一个+ 4R1 ^ 4 ^ 2 / d + 4 ^ FR2 F + 4 ^ fr2d + 4 ^ FR2 F + 4 ^ fr2d + 4 ^ FR2 F + 4 ^ fr2d + 4 ^ FR2 F + 4 ^ FR2 ＆GT; d + 4 ^ FR2 F + 4 ^ fr2d + 4 ^ FR2 F + 4 ^ FR2 ＆LT; F + 4 ^ G + R2 F + 4 ^ fr2f + 4 ^ G + R2 F + 4 ^ fr2f + 4 ^ G + R2 F + 4 ^ fr2f + 4 ^ G + R2 F + 4 ^ fr2f + 4 ^ G + R2 F + 4 ^ fr2f + 4 ^ G + R2 F + 4 ^ fr2f + 4 ^ G + R2 F + 4 ^ fr2f + 4 ^ G + R2 F + 4 ^ FR2 ＆GT; 一个+ 4 ^ G + R2 F + 1A + 4 ^ G + R2 F + 1 F + 4 ^ FR2 d + 1 F + 4 ^ FR2 d + 2 ^ d + 4 ^ r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1
＃3 @ 10 v210 y10
R1 ^ 1 O3 c8r8d8r8 c8r8c8r8c8r8c8r8c8r8c8r8c8r8c8r8c8r8c8r8c8r8 C8 @ 10d16d16 @ 21 C8 @ 10d16d16 @ 21 C8 @ 10d16d16 @ 21 / C4 @ 10d8 @ 21c8＆LT; B8＆GT; C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8＆LT; B8＆GT; C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8＆LT; B8＆GT; C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8＆LT; B8＆GT; C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8＆LT; B8＆GT; C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8＆LT; B8＆GT; C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8＆LT; B8＆GT; C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8＆LT; B8＆GT; C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8＆LT; B8＆GT; C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8＆LT; B8＆GT; C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8＆LT; B8＆GT; C8 @ 10d8 @ 21c8 C4 @ 10d8 @ 21c8＆LT; B8＆GT; @ 10d16d16d16d16d16r16 C4 @ 10d8 @ 21c8＆LT; B8＆GT; C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8＆LT; B8＆GT; C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8＆LT; B8＆GT; C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8＆LT; B8＆GT; C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8＆LT; B8＆GT; C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8＆LT; B8＆GT; C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8＆LT; B8＆GT; C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8＆LT; B8＆GT; C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8＆LT; B8＆GT; C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8＆LT; B8＆GT; C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8＆LT; B8＆GT; C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8＆LT; B8＆GT; C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8＆LT; B8＆GT; C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8＆LT; B8＆GT; C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8＆LT; B8＆GT; C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8＆LT; B8＆GT; C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8＆LT; B8＆GT; C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8＆LT; B8＆GT; C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8＆LT; B8＆GT; C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8＆LT; B8＆GT; C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8＆LT; B8＆GT; C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8＆LT; B8＆GT; C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8＆LT; B8＆GT; C8 @ 10d8 @ 21c8 C4 @ 10d8 @ 21c8 @ 10b16b16＆GT; c16c16＆LT; b16b16a16a16
＃7 @ 16 v230 y10
L16 $ ED $ EF $ A9 cceeggbbggeeccee ＆LT; BB＆GT; d + d + F + F + A + A + F + F + d + d + LT; BB＆GT; d + d + ＆LT; AA＆GT; cceeggeecc＆LT; AA＆GT; CC ＆LT; G + G + BB＆GT; d + d + + FFD d + LT; BBG + G + BB / r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1
＃5 @ 4 v155 y10
L8 $ ED $ F8 $ 8F O4 r1r1r1r1 d + 4R1 ^ 2 ^ 4 / ＆LT;一个+ 4 ^＆GT; CR2 C + 4 ^ CR2＆LT;一个+ 4 ^＆GT; CR2 C + 4 ^ CR2＆LT;一个+ 4 ^＆GT; CR2 C + 4 ^ CR2＆LT;一个+ 4 ^＆GT; CR2 C + 4 ^ CR2 一个+ 4 ^＆GT; CR2 C + 4 ^ CR2 ＆LT;一个+ 4 ^＆GT; CR2 C + 4 ^ C r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1 R2 F + 4 ^ FR2 d + 1F + 4 ^ FR2 d + 1 C + 4 ^ CR2 ＆LT; A + 1 ＆以及c + 4 ^ CR2 ＆LT; A + 2 ^一个+ 4 ^ r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1
答案 0 :(得分：2)
您可以使用Smith-Waterman算法进行局部对齐，将字符串与自身进行比较。
http://en.wikipedia.org/wiki/Smith-Waterman_algorithm
编辑：要使算法适应自对齐，您需要将对角线中的值强制为零 - 也就是说，惩罚将整个字符串与其自身对齐的简单解决方案。然后会弹出“第二好”的对齐方式。这将是最长的两个匹配子串。重复相同的事情，找到逐渐缩短的匹配子串。
答案 1 :(得分：2)
不确定这是否是您要找的。
我把字符串“testtesttesttest4notaduped + c + d + f + d + c + d + f + d + c + d + f + d + c + d + f + testtesttest”并将其转换为“[test] 4 4notadupe [d + c + d + f +] 4 [test] 3“
我确信有人会提出更有效的解决方案，因为在处理完整文件时它会有点慢。我期待着其他答案。
string stringValue = "testtesttesttest4notaduped+c+d+f+d+c+d+f+d+c+d+f+d+c+d+f+testtesttest";
for(int i = 0; i < stringValue.Length; i++)
{
for (int k = 1; (k*2) + i <= stringValue.Length; k++)
{
int count = 1;
string compare1 = stringValue.Substring(i,k);
string compare2 = stringValue.Substring(i + k, k);
//Count if and how many duplicates
while (compare1 == compare2)
{
count++;
k += compare1.Length;
if (i + k + compare1.Length > stringValue.Length)
break;
compare2 = stringValue.Substring(i + k, compare1.Length);
}
if (count > 1)
{
//New code. Added a space to the end to avoid [test]4
//turning using an invalid number ie: [test]44.
string addString = "[" + compare1 + "]" + count + " ";
//Only add code if we are saving space
if (addString.Length < compare1.Length * count)
{
stringValue = stringValue.Remove(i, count * compare1.Length);
stringValue = stringValue.Insert(i, addString);
i = i + addString.Length - 1;
}
break;
}
}
}
答案 2 :(得分：1)
LZW可以提供帮助：它使用前缀字典来搜索重复模式，并使用对先前条目的引用替换此类数据。我认为根据您的需求调整它应该不难。
答案 3 :(得分：0)
为什么不使用System.IO.Compression？