本文目錄一覽:
- 1、JAVA鬥地主如何實現發牌和出牌功能
- 2、java 52張 撲克牌 問題
- 3、撲克牌實現洗牌發牌,顯示各家手上的牌並統計各家的牌中有幾組對子,幾組三條(例如三個A),幾組同花
- 4、java52張撲克牌問題
- 5、用java程序編寫一個發牌的程序。。。跪求了。。。。
JAVA鬥地主如何實現發牌和出牌功能
用隨機的方式生成54張牌(即數字或者其他方式),按順時針方向,依次給每個人一個數字。直到剩最後三張。
出牌:先判斷出牌方選中的牌是否符合規則。是的話。則該人的數值列表中去掉相應的牌。桌面上顯示它出示的牌。出完牌還得判斷遊戲是否結束。如果他的牌下完了。遊戲就結束了
java 52張 撲克牌 問題
識別類:
一張牌:public class Card{};
一副牌:public class Poker{};
帶main方法的類用於實現洗牌和切牌:public class Shuffle{};
識別屬性:
class Card:
一張牌應該具有花色和點數這兩個屬性:
private color; //花色
private value; //點數
class Poker:
一副牌有54張牌,四種花色,每種花色都有2~A的點數:
static Card[] cards=new Card[54];// 每副撲克都有54張牌;
static String[] colors={“紅桃”,”黑桃”,”方塊”,”梅花”}; //每副牌都有四種花色;
static String values[]={“2″,”3″,”4″,”5″,”6″,”7″,”8″,”9″,”10″,”J”,”Q”,”K”,”A”};//每個花
色的點數
定義類:
public class Card{
String color;
String value;
//定義一個構造函數用於初始化點數2~A的牌
public Card(String color,String value){
this.color = color;
this.value = value;
}
//定義一個構造函數用於初始化大王和小王
public Card(String value){
this.color=””; //如果不給color賦值的話,洗牌和切牌後的大小王前面會顯示“null”
this.value = value;
}
// 取一張牌的花色
public String getcolor() {
return color;
}
// 取一張牌的值
public String getvalue() {
return value;
}
}
public class Poker{
static Card[] cards=new Card[54];
static String[] colors={“紅桃”,”黑桃”,”方塊”,”梅花”};
static String values[]={“2″,”3″,”4″,”5″,”6″,”7″,”8″,”9″,”10″,”J”,”Q”,”K”,”A”};
//構造方法Poker()用於初始化這副撲克
public Poker() {
for (int i = 0; i colors.length; i++) {
for (int j = 0; j values.length; j++) {
cards[i * 13 + j] = new Card(colors[i], values[j]);
}
}
cards[52] = new Card(“小王”);
cards[53] = new Card(“大王”);
}
//方法getCard()用於獲取一張牌的花色和點數
public Card[] getCard() {
return Poker.cards;
}
//方法Show()用於顯示一副新的撲克
public void Show() {
……………………
}
public class Shuffle {
static Poker poker = new Poker();
static Card[] cards = new Card[54];
//方法random用於返回一個不重複的數組
public static int[] random(int length) {
int[] a = new int[length];
for (int i = 0; i length; i++) {
a[i] = (int) (Math.random() * length);//產生一個0到length的隨機數
for (int j = 0; j i; j++) {
if (a[j] == a[i]) {//比較是否相等,如果相等則重新產生
i–;
break;
}
}
}
return a;
}
//方法xipai()用於將所得的隨機數組的值賦值給牌的下標
public static void xipai() {
int[] b = new int[54];
b = random(54);//調用random方法產生一個長度為54的不重複的隨機數組
cards = poker.getCard();
Card temp = null;
for(int i = 0; i b.length; i++){ //開始遍曆數組實現重新排列
int k = b[i];
temp = cards[k];
cards[i] = cards[k];
cards[k] = temp;}
//輸出洗牌後的撲克
……………………
……………………
}
//方法qiepai()實現撲克的切牌操作
public static void qiepai(int length) {
int number = (int) (Math.random() * length);//產生一個隨機數
System.out.println(“隨機數為:” + number);//顯示隨機數是多少
Card[] temp = new Card[length];
int start = number;//設置切牌後的起點位置
for (int i = 0; i length – number; i++) {
temp[i] = cards[start++];//將切點後面的牌暫時放入temp數組中
}
for (int j = 0; j number; j++) {
temp[length – number + j] = cards[j];//將切點前面的牌加入temp數組中
}
for (int k = 0; k length; k++) {
cards[k] = temp[k];}
//輸出切牌後的撲克
……………………
……………………
}
//main方法
public static void main(String[] args) {
Poker newpoker = new Poker();
newpoker.Show();//調用Show()方法顯示一副新撲克
xipai();//調用xipai()方法顯示洗牌過後的撲克
qiepai(54);//調用qiepai()方法顯示切牌以後的撲克
}
}
參考資料:
撲克牌實現洗牌發牌,顯示各家手上的牌並統計各家的牌中有幾組對子,幾組三條(例如三個A),幾組同花
1500行。。。。搞笑。200行左右。
這個對對子的處理,存在一個問題,就是有一個三條的存在,那麼該算成是幾對呢?
按排列算法的話,是一共有三對,在這裡只算出了兩對,想要算出三對,會麻煩一些。
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Comparator;
import java.util.List;
import java.util.Random;
public class Cards {
public static void main(String[] args) {
ListCard list = new ArrayListCard();
for (int i = 0; i 4; i++) {
for (int j = 1; j = 13; j++) {
Card card = new Card();
card.type = i;
card.serial = j;
card.execute();
list.add(card);
}
}
//
Card[] west = licensing(list, 13);
Card[] east = licensing(list, 13);
Card[] north = licensing(list, 13);
Card[] south = licensing(list, 13);
System.out.println(“玩家[west]手中的牌是” + Arrays.toString(west));
System.out.println(“玩家[east]手中的牌是” + Arrays.toString(east));
System.out.println(“玩家[north]手中的牌是” + Arrays.toString(north));
System.out.println(“玩家[south]手中的牌是” + Arrays.toString(south));
assort(west, “west”);
assort(east, “east”);
assort(north, “north”);
assort(south, “south”);
}
// 發牌,隨機取出,相當於洗牌
public static Card[] licensing(ListCard list, int len) {
Random random = new Random();
Card[] cards = new Card[len];
int cardIndex = 0;
for (int i = 0; i len; i++) {
int index = random.nextInt(list.size());
cards[cardIndex++] = list.remove(index);
}
// 排序
Arrays.sort(cards, new ComparatorCard() {
@Override
public int compare(Card o1, Card o2) {
if (o1.type != o2.type) {
return o1.type – o2.type;
}
return o1.serial – o2.serial;
}
});
return cards;
}
public static void assort(Card[] cards, String username) {
System.out.print(“玩家[” + username + “]”);
ListCard[] towList = new ArrayListCard[]();
ListCard[] threeList = new ArrayListCard[]();
Card[] tow = null;
Card[] three = null;
for (int i = 0; i cards.length – 1; i++) {
tow = new Card[2];
three = new Card[3];
tow[0] = cards[i];
three[0] = cards[i];
for (int j = i + 1; j cards.length; j++) {
if (cards[i].type == cards[j].type) {
continue;
}
if (cards[i].serial == cards[j].serial) {
tow[1] = cards[j];
towList.add(tow);
three[1] = cards[j];
for (int k = 0; k cards.length; k++) {
if (three[0].type == cards[k].type
|| three[1].type == cards[k].type) {
continue;
}
if (three[1].serial == cards[k].serial) {
three[2] = cards[k];
threeList.add(three);
}
}
break;
}
}
}
ListCard[] flower = new ArrayListCard[]();
// 三張連續的即為同花
for (int i = 0; i cards.length – 2; i++) {
three = new Card[3];
three[0] = cards[i];
three[1] = cards[i + 1];
three[2] = cards[i + 2];
if (three[0].type != three[1].type
|| three[2].type != three[1].type) {
continue;
}
if (three[1].serial – three[0].serial != 1
|| three[2].serial – three[1].serial != 1) {
continue;
}
flower.add(three);
}
System.out.println();
System.out.print(“\t有對子” + towList.size() + “組”);
outList(towList);
System.out.println();
System.out.print(“\t有三條” + threeList.size() + “組”);
outList(threeList);
System.out.println();
System.out.print(“\t有同花” + threeList.size() + “組”);
outList(flower);
System.out.println();
}
public static void outList(ListCard[] list) {
for (int i = 0; i list.size(); i++) {
Card[] cards = list.get(i);
System.out.print(Arrays.toString(cards));
}
}
}
class Card {
String[] types = { “黑桃”, “紅桃”, “梅花”, “方塊” };
int type = 0;
int serial = 0;
String name = null;
public void execute() {
name = types[type];
switch (serial) {
case 1:
name += “A”;
break;
case 11:
name += “J”;
break;
case 12:
name += “Q”;
break;
case 13:
name += “K”;
break;
default:
name += serial;
break;
}
}
public String toString() {
return name;
}
}
java52張撲克牌問題
使用JAVA編寫以下程序:
1、請定義一個名為Card的撲克牌類,該類有兩個private訪問權限的字符串變量face和suit;分別描述一張牌的牌面值(如:A、K、Q、J、10、9、……3、2等)和花色(如:“黑桃”、“紅桃”、“梅花”、“方塊”);定義Card類中的public訪問權限的構造方法,為類中的變量賦值;定義protected訪問權限的方法getFace(),得到撲克牌的牌面值;定義protected訪問權限的方法getSuit(),得到撲克牌的花色;定義方法toString(),返回表示撲克牌的花色和牌面值字符串(如“紅桃A”、“梅花10”等)。
2、若應用程序的main方法中,定義字符串數組f和s;分別表示撲克牌的牌面值和花色;定義52個元素的Card類型數組deck,用來存放4個花色的52張牌。如下所示。
String f[] = {“A”,”2″,”3″,”4″,”5″,”6″,”7″,”8″,”9″,”10″,”J”,”Q”,”K”};
String s[] = {“黑桃”,”紅桃”,”梅花”,”方塊”};
Card deck = new Card[52];
(1)使用Card類的構造方法給deck數組的52張牌賦值,要求數組中先存放黑桃花色的A、2、3、……、K;然後是紅桃花色的A、2、3、……、K;梅花花色的A、2、3、……、K;方塊花色的A、2、3、……、K。請寫出實現上述功能的程序段。
(2)請編寫模擬洗牌的完整程序,即把數組deck中的撲克牌隨機打亂存放順序。
原題如下:
引用內容
private String face; // A, K, Q, J, 10, … 2
private String suit; // Spade, Heart, Club, Diamond
public Card(String suit, String face) {
this.face = face;
this.suit = suit;
}
protected String getFace() {
return face;
}
protected String getSuit() {
return suit;
}
public String toString() {
return suit + ” ” + face;
}
解題,完成程序解:
程序代碼
package poker;
public class Card {
private String face; // A, K, Q, J, 10, … 2
private String suit; // Spade, Heart, Club, Diamond
public Card(String suit, String face) {
this.face = face;
this.suit = suit;
}
protected String getFace() {
return face;
}
protected String getSuit() {
return suit;
}
public String toString() {
return suit + ” ” + face;
}
public static void shuffle(Card[] deck, int startIndex, int size,
int splitIndex) { //shuffle洗牌
System.out.println(“startIndex:” + startIndex);
if (splitIndex * 2 size) { //if rand is 30
Card.swap(deck, startIndex, splitIndex, size – splitIndex); //[52]、0、30、22
shuffle(deck, size – splitIndex, splitIndex, size – splitIndex); //[52]、22、30、22
} else if (splitIndex * 2 size) { //if rnd is 10
Card.swap(deck, startIndex, size – splitIndex, splitIndex); //[52]、0、42、10
shuffle(deck, startIndex, size – splitIndex, splitIndex); //[52]、0、42、10
} else {
Card.swap(deck, startIndex, splitIndex, splitIndex);
}
}
public static void swap(Card[] deck, int srcIndex, int dstIndex, int size) { //交換
//SRC是在本位置顯示:source的縮寫,源的意思 HREF是點擊後連接的目標:HyperlinkReference,超鏈接引用
String face = “”;
String suit = “”;
for (int i = 0; i size; i++) {
face = deck[srcIndex + i].face;
suit = deck[srcIndex + i].suit;
deck[srcIndex + i].face = deck[dstIndex + i].face;
deck[srcIndex + i].suit = deck[dstIndex + i].suit;
deck[dstIndex + i].face = face;
deck[dstIndex + i].suit = suit;
}
}
/**
* @param args
*/
public static void main(String[] args) {
//第一問解答
Card[] deck = new Card[52];
String f[] = {“A”, “2”, “3”, “4”, “5”, “6”, “7”, “8”, “9”, “10”, “J”,
“Q”, “K”};
String s[] = {“Spade”, “Heart”, “Club”, “Diamond”};
for (int i = 0; i s.length; i++) {
for (int j = 0; j f.length; j++) {
deck[i * 13 + j] = new Card(s[i], f[j]); //依次存入數組之中,每13張一回合
}
}
//第二問解答
int rnd = 0;
int numOfShuffle = 10;
for (int i = 0; i numOfShuffle; i++) {
rnd = (int) Math.abs(Math.random() * 52); //Math.random()返回大於或等於 0.0 但小於 1.0 的偽隨機 double 值。
Card.shuffle(deck, 0, deck.length, rnd);
}
// Test
/*
for (int i=0; ideck.length; i++)
{
System.out.println(deck[i]);
} */
}
}
用java程序編寫一個發牌的程序。。。跪求了。。。。
1、設計二維數組a[13][4],13代表2,3,4,5,…,K,A , 4代表紅黑方梅 13*4=52
2、這樣a[0][0]就是紅桃2,a[12][3]就是梅花A
按照這麼來隨機發牌就很簡單了,直接隨機個一個a[i][j]就可以了。i和j都分別在13和4以內
思路明確了,代碼還是自己寫比較好
原創文章,作者:NRMY,如若轉載,請註明出處:https://www.506064.com/zh-hant/n/133198.html