c# - 是否有可能找出字符串列表中的公共(public)部分

Question

我正在努力找出字符串列表中的公共(public)字符串部分。如果我们取一个样本数据集

private readonly List<string> Xpath = new List<string>()
{   
    "BODY>MAIN:nth-of-type(1)>DIV>SECTION>DIV>SECTION>DIV>DIV:nth-of-type(1)>DIV>DIV:nth-of-type(3)>DIV>ARTICLE>DIV>DIV>DIV>SECTION:nth-of-type(1)>H2:nth-of-type(1)",
    "BODY>MAIN:nth-of-type(1)>DIV>SECTION>DIV>SECTION>DIV>DIV:nth-of-type(1)>DIV>DIV:nth-of-type(3)>DIV>ARTICLE>DIV>DIV>DIV>SECTION:nth-of-type(2)>H2:nth-of-type(1)",
    "BODY>MAIN:nth-of-type(1)>DIV>SECTION>DIV>SECTION>DIV>DIV:nth-of-type(1)>DIV>DIV:nth-of-type(3)>DIV>ARTICLE>DIV>DIV>DIV>SECTION:nth-of-type(3)>H2:nth-of-type(1)",
    "BODY>MAIN:nth-of-type(1)>DIV>SECTION>DIV>SECTION>DIV>DIV:nth-of-type(1)>DIV>DIV:nth-of-type(3)>DIV>ARTICLE>DIV>DIV>DIV>SECTION:nth-of-type(4)>H2:nth-of-type(1)",
    "BODY>MAIN:nth-of-type(1)>DIV>SECTION>DIV>SECTION>DIV>DIV:nth-of-type(1)>DIV>DIV:nth-of-type(3)>DIV>ARTICLE>DIV>DIV>DIV>SECTION:nth-of-type(5)>H2:nth-of-type(1)",
    "BODY>MAIN:nth-of-type(1)>DIV>SECTION>DIV>SECTION>DIV>DIV:nth-of-type(1)>DIV>DIV:nth-of-type(3)>DIV>ARTICLE>DIV>DIV>DIV>SECTION:nth-of-type(6)>H2:nth-of-type(1)",
    "BODY>MAIN:nth-of-type(1)>DIV>SECTION>DIV>SECTION>DIV>DIV:nth-of-type(1)>DIV>DIV:nth-of-type(3)>DIV>ARTICLE>DIV>DIV>DIV>SECTION:nth-of-type(7)>H2:nth-of-type(1)",
    "BODY>MAIN:nth-of-type(1)>DIV>SECTION>DIV>SECTION>DIV>DIV:nth-of-type(1)>DIV>DIV:nth-of-type(3)>DIV>ARTICLE>DIV>DIV>DIV>SECTION:nth-of-type(8)>H2:nth-of-type(1)",
    "BODY>MAIN:nth-of-type(1)>DIV>SECTION>DIV>SECTION>DIV>DIV:nth-of-type(1)>DIV>DIV:nth-of-type(3)>DIV>ARTICLE>DIV>DIV>DIV>SECTION:nth-of-type(9)>H2:nth-of-type(1)"
};

由此，我想找出这些 child 与哪些 child 相似。 data 是一个 Xpath 列表。以编程方式我应该能够告诉

预期输出:

BODY>MAIN:nth-of-type(1)>DIV>SECTION>DIV>SECTION>DIV>DIV:nth-of-type(1)>DIV>DIV:nth-of-type(3)>DIV>ARTICLE>DIV>DIV>DIV

为了得到这个我做了这样的事情。我用 >> 分隔每个项目，然后最初为每个数据集创建一个项目列表。

然后用这个找出什么是独特的元素

private IEnumerable<T> GetCommonItems<T>(IEnumerable<T>[] lists)
{
    HashSet<T> hs = new HashSet<T>(lists.First());
    for (int i = 1; i < lists.Length; i++)
    {
        hs.IntersectWith(lists[i]);
    }
    return hs;
}

能够找出唯一值并再次创建数据集。但是，如果这包含 Ex:- Div 在两个地方，并且它也在每个原始数据集中，那么这个方法将只选择一个 Div。

从那时起我会得到这样的东西:

BODY>MAIN:nth-of-type(1)>DIV>SECTION

但是我需要这个

BODY>MAIN:nth-of-type(1)>DIV>SECTION>DIV>SECTION>DIV>DIV:nth-of-type(1)>DIV>DIV:nth-of-type(3)>DIV>ARTICLE>DIV>DIV>DIV

Answer 1

免责声明:这不是最高效的解决方案，但它有效:)

• 让我们开始用 > 字符分割第一条路径
• 对所有路径做同样的事情

char separator = '>';
IEnumerable<string> firstPathChunks = Xpath[0].Split(separator);
var chunks = Xpath.Select(path => path.Split(separator).ToList()).ToArray();

• 遍历 firstPathChunks
  • 遍历 block
  • 如果有匹配则删除第一个元素
  • 如果所有第一个元素都被删除，则将匹配的前缀附加到 sb

void Process(StringBuilder sb)
{
    foreach (var pathChunk in firstPathChunks)
    {
        foreach (var chunk in chunks)
        {
            if (chunk[0] != pathChunk)
            {
                return;
            }
            chunk.RemoveAt(0);
        }
        sb.Append(pathChunk); 
        sb.Append(separator);
    }
}

示例用法

var sb = new StringBuilder();
Process(sb);
Console.WriteLine(sb.ToString());

---

输出

BODY>MAIN:nth-of-type(1)>DIV>SECTION>DIV>SECTION>DIV>DIV:nth-of-type(1)>DIV>DIV:nth-of-type(3)>DIV>ARTICLE>DIV>DIV>DIV>