找到单词的最短连续子序列 [英] Find the shortest continuous subsequence of words

问题描述

您好，



请给我一系列单词S和一系列单词T.如何找到单词中最短的连续子序列S使得T中的单词按顺序出现（在c＃中）

谢谢。



我目前的源代码是：

Hello,

Please if I am given a sequence of words S, and a sequence of words T. How do I find the shortest continuous subsequence of words in S such that the words in T appear in that order (in c#)
Thank you.

My current source code is:

public  bool ContainsSubequence<t>(this IEnumerable<t>parent, IEnumerable<t>target)
   {
       bool foundOneMatch = false;
       var enumeratedTarget = target.ToList();
       int enumPos = 0;

       using (IEnumerator<t>parentEnum = parent.GetEnumerator())
       {
           while (parentEnum.MoveNext())
           {
               if (enumeratedTarget[enumPos].Equals(parentEnum.Current))
               {
                   // Match, so move the target enum forward
                   foundOneMatch = true;
                   if (enumPos == enumeratedTarget.Count - 1)
                   {
                       // We went through the entire target, so we have a match
                       return true;
                   }

                   enumPos++;
               }
               else if (foundOneMatch)
               {
                   foundOneMatch = false;
                   enumPos = 0;

                   if (enumeratedTarget[enumPos].Equals(parentEnum.Current))
                   {
                       foundOneMatch = true;
                       enumPos++;
                   }
               }
           }

           return false;
       }
   }

但它无效

but its not working

推荐答案

我认为一种相当强力的方法会在这里取得成功。

这是在 T 项目（find） > S ，然后在下面的单词中找到第一个出现的第二个 T 项目（他们）等等，直到找到最后一个 T 项目。将其标记为最佳当前消息。继续这样的过程，适当地更新最佳当前解决方案。最终最佳当前解决方案（可能是空的）就是你的结果。

I suppose a rather 'brute force' approach would succeed here.
That is find the first occurrence of the first T item ("find") in S, then find the first occurrence of the second T item ("them") in the following words and so on, until last T item is found. Mark it as 'best current soution'. Continue such a process, updating 'best current solution' appropriately. Eventually the 'best current solution' (possibly empty) is your result.

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亲自尝试，你可能会发现它不是和你想的一样困难！



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