重复的文本查找
我的主要问题是试图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。 PS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，我为拉伸页面道歉：＃0 @ 16 v225 y10 w250 t76
l16 $ ED $ EF $ A9 p20,20> ecegb> d <bgbgecgec d + d + f + a +> c + <a + f + a + f + d + f + d + c cegbgegec ec d + c + d + f + d + c + d + f + d + c + d + f + d + c + d + f + r1 ^ 1
/ l8 r1r1r1r1 f + f + g + cg + r4 a + c + a + g + cg + r4f + f + g + cg + r4 a + c + a + g + cg + r4f + f + g + cg + r4 a + c + a + g + cg + r4 f + f + g + cg + r4 a + c + a + g + r4g + 16f16c + a + 2 ^ g + f + g + 4 f + ff + 4fd + f4 d + c + d + 4c + c c4d + c + d + 4g + 4a + 4 r1 ^ 2 ^ 4 ^ a + 2 ^ g + f + g + 4 f + ff + 4fd + f4 d + c + d + 4c + c c4d + 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 a + 4r1 ^ 4 ^ 2 / d + 4 ^ fr2 f + 4 ^ fr2d + 4 ^ fr2 f + 4 ^ fr2d + 4 ^ fr2 f + 4 ^ fr2d + 4 ^ fr2 f + 4 ^ fr2> d + 4 ^ fr2 f + 4 ^ fr2d + 4 ^ fr2 f + 4 ^ fr2 a + 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 c8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8 c8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8 C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8 C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8 C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8 C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8 C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8 C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8 C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8 C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8 c8 @ 10d8 @ 21c8 c4 @ 10d8 @ 21c8 @ 10d16d16d16d16d16r16 c4 @ 10d8 @ 21c8 c8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8 c8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8 c8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8 C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8 C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8 C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8 C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8 C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8 C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8 C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8 C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8 C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8 C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8 C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8 C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8 C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8 C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8 C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8 C8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8 c8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8 c8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8 c8 @ 10d8 @ 21c8c4 @ 10d8 @ 21c8 c8 @ 10d8 @ 21c8 c4 @ 10d8 @ 21c8 @ 10b16b16> c16c16 <b16b16a16a16
＃7 @ 16 v230 y10
l16 $ ED $ EF $ A9 cceeggbbggeeccee d + d + f + f + a + a + f + f + d + d + d + d + cceeggeecc cc d + d + + FFD + d <BBG + G + BB / r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1
＃5 @ 4 v155 y10
l8 $ ED $ F8 $ 8F o4 r1r1r1r1 d + 4r1 ^ 2 ^ 4 / cr2 c + 4 ^ cr2 cr2 c + 4 ^ cr2 cr2 c + 4 ^ cr2 cr2 c + 4 ^ cr2 a + 4 ^> cr2 c + 4 ^ cr2 cr2 c + 4 ^ c r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1 r2 f + 4 ^ fr2 d + 1f + 4 ^ FR2 d + 1个C + 4 ^ CR2 C + 4 ^ CR2 <A + 2 ^一个+ 4 ^ r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1
不确定这是否是您正在寻找的。
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 [测试] 3“
我确信有人会提出一个更有效的解决方案，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; } } }
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。
https://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自身完全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。 重复相同的事情以找到逐渐缩短的匹配子串。
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来替换此类数据。 我认为根据您的需求调整它应该不难。
为什么不直接使用System.IO.Compression ？
上述就是C#学习教程：重复的文本查找分享的全部内容，如果对大家有所用处且需要了解更多关于C#学习教程，希望大家多多关注---计算机技术网(www.ctvol.com)!
本文来自网络收集，不代表计算机技术网立场，如涉及侵权请联系管理员删除。
ctvol管理联系方式QQ:251552304
本文章地址：https://www.ctvol.com/cdevelopment/942481.html