我决定自己尝试一下,但最终我没有使用正则表达式——我想也许由于所需搜索的简单性,正则表达式会增加比它们节省的复杂性更多的复杂性。不过,我使用了与另一个答案类似的逻辑:计算每个数字的数量并以此为基础进行所有评分:
enum RollScoreType
{
HighDie,
Pair,
TwoPair,
ThreeOfAKind,
SmallStright,
PairSmallStriaght,
FullHouse,
LargeStraight,
FourOfAKind,
FiveOfAKind
}
struct RollScore
{
public RollScoreType type;
public byte highestDie;
public byte nextHighestDie;
public RollScore(RollScoreType type, byte highest, byte next)
{
this.type = type;
this.highestDie = highest;
this.nextHighestDie = next;
}
public override string ToString()
{
return string.Format("{0} {1} {2}", type, highestDie, nextHighestDie);
}
}
static RollScore GetDiceScore(string input)
{
char[] dice = input.ToCharArray();
byte[] diceCount = new byte[6];
for (int i = 0; i < dice.Length; i++)
diceCount[int.Parse(dice[i].ToString())-1]++;
if (Array.IndexOf(diceCount, (byte)5) >= 0)
return new RollScore(RollScoreType.FiveOfAKind, (byte)(Array.IndexOf(diceCount, (byte)5) + 1), 0);
else if (Array.IndexOf(diceCount, (byte)4) >= 0)
return new RollScore(RollScoreType.FourOfAKind, (byte)(Array.IndexOf(diceCount, (byte)4) + 1), (byte)(Array.IndexOf(diceCount, (byte)1) + 1));
else if (Array.IndexOf(diceCount, (byte)3) >= 0)
{
byte three = (byte)(Array.IndexOf(diceCount, (byte)3) + 1);
if (Array.IndexOf(diceCount, (byte)2) >= 0)
{
byte pair = (byte)(Array.IndexOf(diceCount, (byte)2) + 1);
return new RollScore(RollScoreType.FullHouse, Math.Max(pair, three), Math.Min(pair, three));
}
else
return new RollScore(RollScoreType.ThreeOfAKind, three, (byte)(Array.LastIndexOf(diceCount, (byte)1) + 1));
}
else if (Array.IndexOf(diceCount, (byte)2) >= 0)
{
byte pair = (byte)(Array.IndexOf(diceCount, (byte)2) + 1);
byte highPair = (byte)(Array.LastIndexOf(diceCount, (byte)2) + 1);
if (highPair != pair)
return new RollScore(RollScoreType.TwoPair, highPair, pair);
else
{
byte lowMissingDie = (byte)Array.IndexOf(diceCount, (byte)0);
byte highMissingDie = (byte)Array.LastIndexOf(diceCount, (byte)0);
switch (lowMissingDie)
{
case 0:
if (highMissingDie == 5)
return new RollScore(RollScoreType.PairSmallStriaght, 5, 4);
if (highMissingDie == 1)
return new RollScore(RollScoreType.PairSmallStriaght, 6, 5);
break;
case 4:
return new RollScore(RollScoreType.PairSmallStriaght, 4, 3);
}
return new RollScore(RollScoreType.Pair, pair, (byte)(Array.LastIndexOf(diceCount, (byte)1) + 1));
}
}
byte missingDie = (byte)Array.IndexOf(diceCount, (byte)0);
switch(missingDie)
{
case 0:
return new RollScore(RollScoreType.LargeStraight, 6, 5);
case 1:
return new RollScore(RollScoreType.SmallStright, 6, 5);
case 4:
return new RollScore(RollScoreType.SmallStright, 4, 3);
case 5:
return new RollScore(RollScoreType.LargeStraight, 5, 4);
default:
return new RollScore(RollScoreType.HighDie, 6, (byte)(Array.LastIndexOf(diceCount, (byte)1, 3) + 1));
}
}
令我惊讶的是,我发现小顺子和大顺子的概率在 5 骰子掷骰中是相等的。真的吗!?
编辑:固定;我看到当我包括包含一对的小顺子时,小顺子的概率显着上升。
当我想到它时,对子和小顺子可能应该使用对子作为最高骰子,将顺子中的最高点数作为下一个最高点(为了[正确比较两个都是对子和小顺子的掷骰子) )。如果是这样,我将用以下代码替换处理 PairSmallStraight 的代码块:
switch (lowMissingDie)
{
case 0:
if (highMissingDie == 5)
return new RollScore(RollScoreType.PairSmallStriaght, pair, 5);
if (highMissingDie == 1)
return new RollScore(RollScoreType.PairSmallStriaght, pair, 6);
break;
case 4:
return new RollScore(RollScoreType.PairSmallStriaght, pair, 4);
}