正如我所见，您需要使用 3 张牌扑克或 3 张 patti 或同花游戏来构建扑克牌游戏系统，
让我们识别所有涉及的对象
-
卡片 - 有 2 个属性颜色和基数，因此类卡片将是
class Card {
char color;
short number;
public Card(char color, short number) {
this.color = color;
this.number = number;
}
public char getColor() {
return color;
}
public void setColor(char color) {
this.color = color;
}
public short getNumber() {
return number;
}
public void setNumber(short number) {
this.number = number;
}
}
-
牌组 - 牌组有 52 张牌，因此
class Deck {
char[] deckColors = {'♦', '♠', '♣', '♥'};
short[] cardNum = {(short) 'A', (short) '2', (short) '3', (short) '4', (short) '5', (short) '6', (short) '7',
(short) '8', (short) '9', (short) 'T', (short) 'J', (short) 'Q', (short) 'K'};
int cardCount;
public Card[] getShuffledDeck() {
Random r = new Random();
Card[] deckCards = new Card[(deckColors.length * cardNum.length)];
int cnt = 0;
for (char c : deckColors) {
for (short s : cardNum) {
deckCards[cnt++] = new Card(c, s);
}
}
Card[] shuffledDeck = new Card[deckCards.length];
int addedCount = 0;
while (addedCount < deckCards.length) {
int tInt = r.nextInt((deckCards.length));
Card c = deckCards[tInt];
if (c != null) {
shuffledDeck[addedCount++] = c;
deckCards[tInt] = null;
} else {
}
}
return shuffledDeck;
}
}
-
Hand - 一组 3 张牌，但最好使用 getHandRanking() 之类的方法创建一个类，该方法将实际计算发牌手牌的强度
class Hand {
Card[] cards;
int handRank;
public Hand(Card[] cards) {
this.cards = new Card[3];
//sort all cards
if (cards[0].getNumber() > cards[1].getNumber()) {
if (cards[0].getNumber() > cards[2].getNumber()) {
//0 index is highest card
this.cards[2] = cards[0];
if (cards[2].getNumber() > cards[1].getNumber()) {
//2 is second highest
this.cards[1] = cards[2];
this.cards[0] = cards[1];
} else {
//1 is second highest
this.cards[1] = cards[1];
this.cards[0] = cards[2];
}
} else {
//2 index is highest
this.cards[2] = cards[2];
if (cards[0].getNumber() > cards[1].getNumber()) {
//0 is second highest
this.cards[1] = cards[0];
this.cards[0] = cards[1];
} else {
//1 is second highest
this.cards[1] = cards[1];
this.cards[0] = cards[0];
}
}
} else {
if (cards[1].getNumber() > cards[2].getNumber()) {
//1 index is highest card
this.cards[2] = cards[1];
if (cards[2].getNumber() > cards[0].getNumber()) {
//2 is second highest
this.cards[1] = cards[2];
this.cards[0] = cards[0];
} else {
//0 is second highest
this.cards[1] = cards[0];
this.cards[0] = cards[2];
}
} else {
//2 index is highest
this.cards[2] = cards[2];
if (cards[0].getNumber() > cards[1].getNumber()) {
//0 is second highest
this.cards[1] = cards[0];
this.cards[0] = cards[1];
} else {
//1 is second highest
this.cards[1] = cards[1];
this.cards[0] = cards[0];
}
}
}
}
public int getHandRank() {
return handRank > 0 ? handRank : calculateHandRank();
}
public int calculateHandRank() {
//assuming 3 cards dealt
//Trial - ColorSeq - Seq - Color - Pair
int[] powerOf2s = {1, 2, 4, 8, 16};
return ((cards[0].getNumber() == cards[1].getNumber() && cards[1].getNumber() == cards[2].getNumber()) ? 1 : 0) * powerOf2s[4]
+ (((cards[2].getNumber() - 1 == cards[1].getNumber() && cards[1].getNumber() - 1 == cards[0].getNumber()) && (cards[2].getColor() == cards[1].getColor() && cards[1].getColor() == cards[0].getColor())) ? 1 : 0) * powerOf2s[3]
+ ((cards[2].getNumber() - 1 == cards[1].getNumber() && cards[1].getNumber() - 1 == cards[0].getNumber()) ? 1 : 0) * powerOf2s[2]
+ (((cards[2].getColor() == cards[1].getColor() && cards[1].getColor() == cards[0].getColor())) ? 1 : 0) * powerOf2s[1]
+ ((cards[0].getNumber() == cards[1].getNumber() || cards[1].getNumber() == cards[2].getNumber() || cards[0].getNumber() == cards[2].getNumber()) ? 1 : 0) * powerOf2s[0];
}
}
现在您只需要查看所有玩家手牌中哪个玩家的手牌等级最高，如果两个玩家的手牌等级相同，则按 Hand.cards[2],[1] 的顺序查看谁的牌高。 [0]。
如果需要任何解释，请告诉我。
算法可以大大改进，给出的代码示例只是展示思考过程。