这是我从the Wikipedia entry on Boyer-Moore的Java版本转换而来的C#版本.

public sealed class BoyerMoore
{
    readonly byte[] needle;
    readonly int[] charTable;
    readonly int[] offsetTable;
 
    public BoyerMoore(byte[] needle)
    {
        this.needle = needle;
        this.charTable = makeByteTable(needle);
        this.offsetTable = makeOffsetTable(needle);
    }
 
    public IEnumerable<int> Search(byte[] haystack)
    {
        if (needle.Length == 0)
            yield break;
 
        for (int i = needle.Length - 1; i < haystack.Length;)
        {
            int j;
 
            for (j = needle.Length - 1; needle[j] == haystack[i]; --i,--j)
            {
                if (j != 0)
                    continue;
 
                yield return i;
                i += needle.Length - 1;
                break;
            }
 
            i += Math.Max(offsetTable[needle.Length - 1 - j],charTable[haystack[i]]);
        }
    }
 
    static int[] makeByteTable(byte[] needle)
    {
        const int ALPHABET_SIZE = 256;
        int[] table = new int[ALPHABET_SIZE];
 
        for (int i = 0; i < table.Length; ++i)
            table[i] = needle.Length;
 
        for (int i = 0; i < needle.Length - 1; ++i)
            table[needle[i]] = needle.Length - 1 - i;
 
        return table;
    }
 
    static int[] makeOffsetTable(byte[] needle)
    {
        int[] table = new int[needle.Length];
        int lastPrefixPosition = needle.Length;
 
        for (int i = needle.Length - 1; i >= 0; --i)
        {
            if (isPrefix(needle,i + 1))
                lastPrefixPosition = i + 1;
 
            table[needle.Length - 1 - i] = lastPrefixPosition - i + needle.Length - 1;
        }
 
        for (int i = 0; i < needle.Length - 1; ++i)
        {
            int slen = suffixLength(needle,i);
            table[slen] = needle.Length - 1 - i + slen;
        }
 
        return table;
    }
 
    static bool isPrefix(byte[] needle,int p)
    {
        for (int i = p,j = 0; i < needle.Length; ++i,++j)
            if (needle[i] != needle[j])
                return false;
 
        return true;
    }
 
    static int suffixLength(byte[] needle,int p)
    {
        int len = 0;
 
        for (int i = p,j = needle.Length - 1; i >= 0 && needle[i] == needle[j]; --i,--j)
            ++len;
 
        return len;
    }
}

这是一些控制台应用测试代码：

public static void Main()
{
    byte[] haystack = new byte[10000];
    byte[] needle = { 0x00,0x69,0x73,0x6F,0x6D };
 
    // Put a few copies of the needle into the haystack.
 
    for (int i = 1000; i <= 9000; i += 1000) 
        Array.Copy(needle,haystack,i,needle.Length);
 
    var searcher = new BoyerMoore(needle);
 
    foreach (int index in searcher.Search(haystack))
        Console.WriteLine(index);
}

注意Search()方法如何返回haystack中针头起始点的所有位置的索引.

如果您只是想要计数,您可以这样做：

int count = new BoyerMoore(needle).Search(haystack).Count();

对于你的第二个问题：我假设你问的是找到最长的重复字节序列？

这是一个更加复杂 – 而且非常不同 – 的问题.如果你想要一个答案,你应该问一个单独的问题,但你应该阅读the Wikipedia entry on the “longest repeated substring problem”.