我敢肯定,這不是最好的方法,根據模式的長度,可能會有運行時間和內存使用情況不起作用,但這裏有一些代碼。
您可以將下面的代碼粘貼到LINQPad並運行它,它應該會產生以下的輸出:
ABCBCABCBCDEEF = (2A(2BC))D(2E)F
ABBABBABBABA = (3A(2B))ABA
ABCDABCDCDCDCD = (2ABCD)(3CD)
正如你所看到的,編碼ABB
爲A(2B)
,而不是ABB
中間的例子中,你將有如果像這樣的單符號序列應該被編碼爲重複符號,或者如果應該使用特定的閾值(如3或更多),那麼您自己做出判斷。
基本上,代碼運行是這樣的:
- 針對序列中的每一個位置,試圖找到最長的匹配(實際上,它沒有,它需要第一2+匹配發現,我因爲我必須現在離開我的計算機幾小時)
- 然後它會嘗試編碼該序列,遞歸地重複該序列,並且吐出一個X * seq類型的對象
- 如果無法找到重複序列,則會在該位置分出單個符號
- 然後,它會跳過它所編碼,並從#1
反正下去,下面的代碼:
void Main()
{
string[] examples = new[]
{
"ABCBCABCBCDEEF",
"ABBABBABBABA",
"ABCDABCDCDCDCD",
};
foreach (string example in examples)
{
StringBuilder sb = new StringBuilder();
foreach (var r in Encode(example))
sb.Append(r.ToString());
Debug.WriteLine(example + " = " + sb.ToString());
}
}
public static IEnumerable<Repeat<T>> Encode<T>(IEnumerable<T> values)
{
return Encode<T>(values, EqualityComparer<T>.Default);
}
public static IEnumerable<Repeat<T>> Encode<T>(IEnumerable<T> values, IEqualityComparer<T> comparer)
{
List<T> sequence = new List<T>(values);
int index = 0;
while (index < sequence.Count)
{
var bestSequence = FindBestSequence<T>(sequence, index, comparer);
if (bestSequence == null || bestSequence.Length < 1)
throw new InvalidOperationException("Unable to find sequence at position " + index);
yield return bestSequence;
index += bestSequence.Length;
}
}
private static Repeat<T> FindBestSequence<T>(IList<T> sequence, int startIndex, IEqualityComparer<T> comparer)
{
int sequenceLength = 1;
while (startIndex + sequenceLength * 2 <= sequence.Count)
{
if (comparer.Equals(sequence[startIndex], sequence[startIndex + sequenceLength]))
{
bool atLeast2Repeats = true;
for (int index = 0; index < sequenceLength; index++)
{
if (!comparer.Equals(sequence[startIndex + index], sequence[startIndex + sequenceLength + index]))
{
atLeast2Repeats = false;
break;
}
}
if (atLeast2Repeats)
{
int count = 2;
while (startIndex + sequenceLength * (count + 1) <= sequence.Count)
{
bool anotherRepeat = true;
for (int index = 0; index < sequenceLength; index++)
{
if (!comparer.Equals(sequence[startIndex + index], sequence[startIndex + sequenceLength * count + index]))
{
anotherRepeat = false;
break;
}
}
if (anotherRepeat)
count++;
else
break;
}
List<T> oneSequence = Enumerable.Range(0, sequenceLength).Select(i => sequence[startIndex + i]).ToList();
var repeatedSequence = Encode<T>(oneSequence, comparer).ToArray();
return new SequenceRepeat<T>(count, repeatedSequence);
}
}
sequenceLength++;
}
// fall back, we could not find anything that repeated at all
return new SingleSymbol<T>(sequence[startIndex]);
}
public abstract class Repeat<T>
{
public int Count { get; private set; }
protected Repeat(int count)
{
Count = count;
}
public abstract int Length
{
get;
}
}
public class SingleSymbol<T> : Repeat<T>
{
public T Value { get; private set; }
public SingleSymbol(T value)
: base(1)
{
Value = value;
}
public override string ToString()
{
return string.Format("{0}", Value);
}
public override int Length
{
get
{
return Count;
}
}
}
public class SequenceRepeat<T> : Repeat<T>
{
public Repeat<T>[] Values { get; private set; }
public SequenceRepeat(int count, Repeat<T>[] values)
: base(count)
{
Values = values;
}
public override string ToString()
{
return string.Format("({0}{1})", Count, string.Join("", Values.Select(v => v.ToString())));
}
public override int Length
{
get
{
int oneLength = 0;
foreach (var value in Values)
oneLength += value.Length;
return Count * oneLength;
}
}
}
public class GroupRepeat<T> : Repeat<T>
{
public Repeat<T> Group { get; private set; }
public GroupRepeat(int count, Repeat<T> group)
: base(count)
{
Group = group;
}
public override string ToString()
{
return string.Format("({0}{1})", Count, Group);
}
public override int Length
{
get
{
return Count * Group.Length;
}
}
}
你能否澄清Huffman或Golomb編碼是否足夠? – Iterator
'ABABABAB'是'4(AB)'還是'2(2(AB))'?或'2(ABAB)'? –
這些模式有多長? –