本文整理汇总了C#中Hand.Remove方法的典型用法代码示例。如果您正苦于以下问题：C# Hand.Remove方法的具体用法？C# Hand.Remove怎么用？C# Hand.Remove使用的例子？那么, 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类Hand的用法示例。
在下文中一共展示了Hand.Remove方法的5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: getKickers
//get all remaining cards, if necessary, to form 5 cards
private static Hand getKickers(Hand hand, Hand specialCards)
{
if (specialCards.Count() == 0)
return specialCards;
for (int i = 0; i < specialCards.Count(); i++)
{
hand.Remove(specialCards.getCard(i));
}
for (int i = 0; i < hand.Count();i++)
{
if (specialCards.Count() >= 5)
break;
specialCards.Add(hand.getCard(i));
specialCards.setValue(hand.getCard(i).getRank());
}
return specialCards;
}示例2: 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;
}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;
}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;
}示例5: 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;
}