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C# Hand.sortByRank方法代码示例

本文整理汇总了C#中Hand.sortByRank方法的典型用法代码示例。如果您正苦于以下问题:C# Hand.sortByRank方法的具体用法?C# Hand.sortByRank怎么用?C# Hand.sortByRank使用的例子?那么恭喜您, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在Hand的用法示例。


在下文中一共展示了Hand.sortByRank方法的15个代码示例,这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞,您的评价将有助于系统推荐出更棒的C#代码示例。

示例1: isStraightFlush

        //use recursion to get rid of pairs, then evaluate straight flush
        public static bool isStraightFlush(Hand hand)
        {
            hand.sortByRank();
            Hand simplifiedhand1, simplifiedhand2; //to be set the same as hand - cards are removed from this hand to evaluate straights separately without the interference of pairs or three-of-a-kind
            for (int i = 0; i <= hand.Count() - 2; i++)
            {
                if (hand.getCard(i) == hand.getCard(i + 1))
                {
                    simplifiedhand1 = new Hand(hand);
                    simplifiedhand1.Remove(i);
                    simplifiedhand2 = new Hand(hand);
                    simplifiedhand2.Remove(i + 1);
                    if (HandCombination.isStraightFlush(simplifiedhand1))
                        return true;
                    if (HandCombination.isStraightFlush(simplifiedhand2))
                        return true;
                }
            }
            for (int i = 0; i <= hand.Count() - 5; i++)
            {
                int currentrank = hand.getCard(i).getRank(), currentsuit = hand.getCard(i).getSuit();
                if (currentrank == hand.getCard(i + 1).getRank() + 1 && currentrank == hand.getCard(i + 2).getRank() + 2 && currentrank == hand.getCard(i + 3).getRank() + 3 && currentrank == hand.getCard(i + 4).getRank() + 4 && currentsuit == hand.getCard(i + 1).getSuit() && currentsuit == hand.getCard(i + 2).getSuit() && currentsuit == hand.getCard(i + 3).getSuit() && currentsuit == hand.getCard(i + 4).getSuit())
                    return true;

            }
            for (int i = 0; i <= hand.Count() - 4; i++)
            {
                int currentrank = hand.getCard(i).getRank(), currentsuit = hand.getCard(i).getSuit();
                if (currentrank == 5 && hand.getCard(i + 1).getRank() == 4 && hand.getCard(i + 2).getRank() == 3 && hand.getCard(i + 3).getRank() == 2 && hand.getCard(0).getRank() == 14 && currentsuit == hand.getCard(i + 1).getSuit() && currentsuit == hand.getCard(i + 2).getSuit() && currentsuit == hand.getCard(i + 3).getSuit() && currentsuit == hand.getCard(0).getSuit())
                    return true;
            }
            return false;
        }
开发者ID:Rodbourn,项目名称:TexasHoldem,代码行数:34,代码来源:HandCombination.cs

示例2: isThreeOfAKind

 public static bool isThreeOfAKind(Hand hand)
 {
     hand.sortByRank();
     for (int i = 0; i <= hand.Count() - 3; i++)
     {
         if (hand.getCard(i) == hand.getCard(i + 1) && hand.getCard(i) == hand.getCard(i + 2))
             return true;
     }
     return false;
 }
开发者ID:Rodbourn,项目名称:TexasHoldem,代码行数:10,代码来源:HandCombination.cs

示例3: isRoyalFlush

 //look for royal flush, removing pair using recursion
 public static bool isRoyalFlush(Hand hand)
 {
     hand.sortByRank();
     Hand simplifiedhand1, simplifiedhand2; //to be set the same as hand - cards are removed from this hand to evaluate straights separately without the interference of pairs or three-of-a-kind
     for (int i = 0; i <= hand.Count() - 2; i++)
     {
         if (hand[i] == hand[i+1])
         {
             simplifiedhand1 = new Hand(hand);
             simplifiedhand1.Remove(i);
             simplifiedhand2 = new Hand(hand);
             simplifiedhand2.Remove(i + 1);
             if (HandCombination.isRoyalFlush(simplifiedhand1))
                 return true;
             if (HandCombination.isRoyalFlush(simplifiedhand2))
                 return true;
         }
     }
     int currentsuit = hand.getCard(0).getSuit();
     if (hand.getCard(0).getRank() == 14 && hand.getCard(1).getRank() == 13 && hand.getCard(2).getRank() == 12 && hand.getCard(3).getRank() == 11 && hand.getCard(4).getRank() == 10 && hand.getCard(1).getSuit() == currentsuit && hand.getCard(2).getSuit() == currentsuit && hand.getCard(3).getSuit() == currentsuit && hand.getCard(4).getSuit() == currentsuit)
         return true;
     else
         return false;
 }
开发者ID:Rodbourn,项目名称:TexasHoldem,代码行数:25,代码来源:HandCombination.cs

示例4: isStraight

 //explanation below
 public static bool isStraight(Hand hand)
 {
     hand.sortByRank();
     if(hand.getCard(0).getRank()==14)
         hand.Add(new Card((int)RANK.ACE,hand.getCard(0).getSuit()));
     int straightCount=1;
     for (int i = 0; i <= hand.Count() - 2; i++)
     {
         //if 5 cards are found to be straights, break out of the loop
         if (straightCount == 5)
             break;
         int currentrank = hand.getCard(i).getRank();
         //if cards suit differ by 1, increment straight
         if (currentrank - hand.getCard(i + 1).getRank() == 1)
             straightCount++;
         //specific condition for 2-A
         else if (currentrank == 2 && hand.getCard(i + 1).getRank() == 14)
             straightCount++;
         //if cards suit differ by more than 1, reset straight to 1
         else if (currentrank - hand.getCard(i + 1).getRank() > 1)
             straightCount = 1;
         //if card suits does not differ, do nothing
     }
     if (hand.getCard(0).getRank() == 14)
         hand.Remove(hand.Count() - 1);
     //depending on the straight count, return true or false
     if (straightCount == 5)
         return true;
     return false;
 }
开发者ID:Rodbourn,项目名称:TexasHoldem,代码行数:31,代码来源:HandCombination.cs

示例5: isFullHouse

 public static bool isFullHouse(Hand hand)
 {
     hand.sortByRank();
     bool threeofakind = false, pair = false;
     int threeofakindRank = 0;
     for (int i = 0; i <= hand.Count() - 3; i++)
     {
         if (hand.getCard(i) == hand.getCard(i + 1) && hand.getCard(i) == hand.getCard(i + 2))
         {
             threeofakind = true;
             threeofakindRank = hand.getCard(i).getRank();
             break;
         }
     }
     for (int i = 0; i <= hand.Count() - 2; i++)
     {
         if (hand.getCard(i) == hand.getCard(i + 1) && hand.getCard(i).getRank() != threeofakindRank)
         {
             pair = true;
             break;
         }
     }
     if (threeofakind == true && pair == true)
         return true;
     else
         return false;
 }
开发者ID:Rodbourn,项目名称:TexasHoldem,代码行数:27,代码来源:HandCombination.cs

示例6: isOnePair

 public static bool isOnePair(Hand hand)
 {
     hand.sortByRank();
     for (int i = 0; i <= hand.Count() - 2; i++)
     {
         if (hand.getCard(i) == hand.getCard(i + 1))
             return true;
     }
     return false;
 }
开发者ID:Rodbourn,项目名称:TexasHoldem,代码行数:10,代码来源:HandCombination.cs

示例7: getTwoPair

 public static Hand getTwoPair(Hand hand)
 {
     hand.sortByRank();
     Hand twopair = new Hand();
     twopair.setValue(3);
     int pairCount = 0;
     for (int i = 0; i <= hand.Count() - 2; i++)
     {
         if (hand.getCard(i) == hand.getCard(i + 1))
         {
             twopair.setValue(hand.getCard(i).getRank());
             twopair.Add(hand.getCard(i));
             twopair.Add(hand.getCard(i+1));
             pairCount++;
             if (pairCount == 2)
                 break;
             i++; //the pair has already been checked, i must be incremented an additional time to avoid using a card in this pair again. This prevents the program from identifying 3 of a kind as 2 pairs.
         }
     }
     if (pairCount == 2)
         return getKickers(hand,twopair);
     else
         twopair.Clear();
         return twopair;
 }
开发者ID:Rodbourn,项目名称:TexasHoldem,代码行数:25,代码来源:HandCombination.cs

示例8: getFullHouse

 public static Hand getFullHouse(Hand hand)
 {
     hand.sortByRank();
     Hand fullhouse = new Hand();
     fullhouse.setValue(7);
     bool threeofakind = false, pair = false;
     int threeofakindRank = 0;
     for (int i = 0; i <= hand.Count() - 3; i++)
     {
         if (hand.getCard(i) == hand.getCard(i + 1) && hand.getCard(i) == hand.getCard(i + 2))
         {
             threeofakind = true;
             threeofakindRank = hand.getCard(i).getRank();
             fullhouse.Add(hand.getCard(i));
             fullhouse.Add(hand.getCard(i + 1));
             fullhouse.Add(hand.getCard(i + 2));
             fullhouse.setValue(hand.getCard(i).getRank());
             break;
         }
     }
     for (int i = 0; i <= hand.Count() - 2; i++)
     {
         if (hand.getCard(i) == hand.getCard(i + 1) && hand.getCard(i).getRank() != threeofakindRank)
         {
             pair = true;
             fullhouse.Add(hand.getCard(i));
             fullhouse.Add(hand.getCard(i + 1));
             fullhouse.setValue(hand.getCard(i).getRank());
             break;
         }
     }
     if (threeofakind == true && pair == true)
         return fullhouse;
     else
     {
         fullhouse.Clear();
         return fullhouse;
     }
 }
开发者ID:Rodbourn,项目名称:TexasHoldem,代码行数:39,代码来源:HandCombination.cs

示例9: getStraightFlush

 //get straight flush using two pointer variable and taking care of all cases
 public static Hand getStraightFlush(Hand hand)
 {
     hand.sortByRank();
     Hand straightflush = new Hand();
     straightflush.setValue(9);
     if (hand.getCard(0).getRank() == 14)
         hand.Add(new Card((int)RANK.ACE, hand.getCard(0).getSuit()));
     //int straightflushCount = 1;
     straightflush.Add(hand.getCard(0));
     int ptr1=0, ptr2=1;
     while (ptr1 < hand.Count() - 2 || ptr2 < hand.Count())
     {
         if (straightflush.Count() >= 5)
             break;
         int rank1=hand.getCard(ptr1).getRank(), rank2=hand.getCard(ptr2).getRank();
         int suit1=hand.getCard(ptr1).getSuit(), suit2=hand.getCard(ptr2).getSuit();
         if (rank1 - rank2 == 1 && suit1 == suit2)
         {
             straightflush.Add(hand.getCard(ptr2));
             ptr1 = ptr2;
             ptr2++;
         }
         else if(rank1==2&&rank2==14&&suit1==suit2)
         {
             straightflush.Add(hand.getCard(ptr2));
             ptr1 = ptr2;
             ptr2++;
         }
         else
         {
             if (rank1 - rank2 <= 1)
                 ptr2++;
             else
             {
                 straightflush.Clear();
                 straightflush.setValue(9);
                 ptr1++;
                 ptr2=ptr1+1;
                 straightflush.Add(hand.getCard(ptr1));
             }
         }
     }
     if (hand.getCard(0).getRank() == 14)
         hand.Remove(hand.Count() - 1);
     straightflush.setValue(straightflush.getCard(0).getRank());
     if (straightflush.Count() < 5)
         straightflush.Clear();
     return straightflush;
 }
开发者ID:Rodbourn,项目名称:TexasHoldem,代码行数:50,代码来源:HandCombination.cs

示例10: getThreeOfAKind

 public static Hand getThreeOfAKind(Hand hand)
 {
     hand.sortByRank();
     Hand threeofakind = new Hand();
     threeofakind.setValue(4);
     for (int i = 0; i <= hand.Count() - 3; i++)
     {
         if (hand.getCard(i) == hand.getCard(i + 1) && hand.getCard(i) == hand.getCard(i + 2))
         {
             threeofakind.setValue(hand.getCard(i).getRank());
             threeofakind.Add(hand.getCard(i));
             threeofakind.Add(hand.getCard(i + 1));
             threeofakind.Add(hand.getCard(i + 2));
             break;
         }
     }
     return getKickers(hand, threeofakind);
 }
开发者ID:Rodbourn,项目名称:TexasHoldem,代码行数:18,代码来源:HandCombination.cs

示例11: getRoyalFlush

 //get royal flush using recursion
 public static Hand getRoyalFlush(Hand hand)
 {
     hand.sortByRank();
     Hand straightflush = new Hand(HandCombination.getStraightFlush(hand));
     straightflush.setValue(10);
     if (straightflush.getCard(0).getRank() == 14)
         return straightflush;
     else
     {
         straightflush.Clear();
         return straightflush;
     }
 }
开发者ID:Rodbourn,项目名称:TexasHoldem,代码行数:14,代码来源:HandCombination.cs

示例12: getOnePair

 public static Hand getOnePair(Hand hand)
 {
     hand.sortByRank();
     Hand onepair = new Hand();
     onepair.setValue(2);
     for (int i = 0; i <= hand.Count() - 2; i++)
     {
         if (hand.getCard(i) == hand.getCard(i + 1))
         {
             onepair.setValue(hand.getCard(i).getRank());
             onepair.Add(hand.getCard(i));
             onepair.Add(hand.getCard(i + 1));
             break;
         }
     }
     return getKickers(hand, onepair);
 }
开发者ID:Rodbourn,项目名称:TexasHoldem,代码行数:17,代码来源:HandCombination.cs

示例13: getHighCard

 //get highest cards after sorting
 public static Hand getHighCard(Hand hand)
 {
     hand.sortByRank();
     Hand highcard = new Hand();
     highcard.setValue(1);
     highcard.Add(hand.getCard(0));
     highcard.setValue(hand.getCard(0).getRank());
     return getKickers(hand, highcard);
 }
开发者ID:Rodbourn,项目名称:TexasHoldem,代码行数:10,代码来源:HandCombination.cs

示例14: isTwoPair

 public static bool isTwoPair(Hand hand)
 {
     hand.sortByRank();
     int pairCount = 0;
     for (int i = 0; i <= hand.Count() - 2; i++)
     {
         if (hand.getCard(i) == hand.getCard(i + 1))
         {
             pairCount++;
             i++; //the pair has already been checked, i must be incremented an additional time to avoid using a card in this pair again. This prevents the program from identifying 3 of a kind as 2 pairs.
         }
     }
     if (pairCount >= 2)
         return true;
     else
         return false;
 }
开发者ID:Rodbourn,项目名称:TexasHoldem,代码行数:17,代码来源:HandCombination.cs

示例15: isFlush

 //use a counter, if a counter reaches five, a flush is detected
 public static bool isFlush(Hand hand)
 {
     hand.sortByRank();
     int diamondCount = 0, clubCount = 0, heartCount = 0, spadeCount = 0;
     for (int i = 0; i < hand.Count(); i++)
     {
         if ((SUIT)hand.getCard(i).getSuit() == SUIT.DIAMONDS)
             diamondCount++;
         else if ((SUIT)hand.getCard(i).getSuit() == SUIT.CLUBS)
             clubCount++;
         else if ((SUIT)hand.getCard(i).getSuit() == SUIT.HEARTS)
             heartCount++;
         else if ((SUIT)hand.getCard(i).getSuit() == SUIT.SPADES)
             spadeCount++;
     }
     if (diamondCount >= 5)
         return true;
     else if (clubCount >= 5)
         return true;
     else if (heartCount >= 5)
         return true;
     else if (spadeCount >= 5)
         return true;
     return false;
 }
开发者ID:Rodbourn,项目名称:TexasHoldem,代码行数:26,代码来源:HandCombination.cs


注:本文中的Hand.sortByRank方法示例由纯净天空整理自Github/MSDocs等开源代码及文档管理平台,相关代码片段筛选自各路编程大神贡献的开源项目,源码版权归原作者所有,传播和使用请参考对应项目的License;未经允许,请勿转载。