這里只介紹普通的麻將胡牌算法，也就是7個(gè)對子或者 1個(gè)對子+3*N; N = 三個(gè)順子或者三個(gè)一樣的 ，其中字牌（東南西北中發(fā)白）不能算順子。

首先對于每張牌 它有牌的的數(shù)字 1-9，牌的種類 （萬條筒以及字）

所以先封裝一個(gè)牌的包裝類。如果要和服務(wù)器交互的話就應(yīng)該給這個(gè)類序列化 即加上System.Serializable，然后再轉(zhuǎn)Json格式，這里只做胡牌的基本判斷。

public class Card{private int number;public int Number{get { return number; }set { number = value; }}private int type;public int Type{get { return type; }set { type = value; }}public Card(int n, int t){number = n;type = t;}}

首先對于一副牌應(yīng)該先判斷是否滿足七對。不滿足則向下判斷。對于手牌會(huì)創(chuàng)建一個(gè)交錯(cuò)數(shù)組來保管。即

int[][] handcards = new int[4][] { new int[] { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 }, new int[] { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 },
? ? ? ? ? ? ? ? new int[] { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 }, new int[] { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 } };

每個(gè)數(shù)組的第一位（handcards[i][0]）用來保存當(dāng)前類型牌的總數(shù)。

對應(yīng)每種類型首先分為是否包含一個(gè)對子或者不包含對子。

所以對于四種類型應(yīng)該是只有一種包含一個(gè)對子其他不包含對子。

代碼如下：

//type為0,1,2,3則為萬條筒字class Program{public void asd(){}static void Main(string[] args){Card c1 = new Card(1, 1);Card c2 = new Card(1, 1);Card c3 = new Card(2, 1);Card c4 = new Card(2, 1);Card c5 = new Card(2, 1);Card c6 = new Card(4, 1);Card c7 = new Card(4, 1);Card c8 = new Card(4, 1);Card c9 = new Card(4, 3);Card c10 = new Card(5, 3);Card c11 = new Card(6, 3);Card c12 = new Card(6, 1);Card c13 = new Card(6, 1);Card c14 = new Card(6, 1);Card[] cs = { c1, c2, c3, c4, c5, c6, c7, c8, c9, c10, c11, c12, c13, c14 };Console.WriteLine(isHu(cs));}public static bool isHu(Card[] cards){int countCards = cards.Length;//判斷是否是七小對if (countCards == 14){int count = 0;for (int i = 0; i < cards.Length; i += 2){if (cards[i].Number == cards[i + 1].Number){count++;}else{break;}}if (count == 7){return true;}}int[][] handcards = new int[4][] { new int[] { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 }, new int[] { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 },new int[] { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 }, new int[] { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 } };for (int i = 0; i < cards.Length; i++){//type為0,1,2,3則為萬條筒字。 [type = 0,1,2,3 , 0]代表每種牌型的總數(shù)量switch (cards[i].Type){case 0: handcards[0][cards[i].Number]++;handcards[0][0]++;break;case 1: handcards[1][cards[i].Number]++;handcards[1][0]++;break;case 2: handcards[2][cards[i].Number]++;handcards[2][0]++;break;case 3: handcards[3][cards[i].Number]++;handcards[3][0]++;break;}}bool isJiang = false; //判斷是否有對子int jiangNumber = -1;for (int i = 0; i < handcards.GetLength(0); i++){if (handcards[i][0] % 3 == 2){if (isJiang){return false;}isJiang = true;jiangNumber = i;}//因?yàn)閷?yīng)四種牌型只能有一種且僅包含一個(gè)對子}//先求沒有將牌的情況判斷其是不是都是由刻子或者砍組成for (int i = 0; i < handcards.GetLength(0); i++){if (i != jiangNumber){if (!(IsKanOrShun(handcards[i], i == 3))){return false;}}}bool success = false;//有將牌的情況下for (int i = 1; i <= 9; i++){if (handcards[jiangNumber][i] >= 2){handcards[jiangNumber][i] -= 2;handcards[jiangNumber][0] -= 2;if (IsKanOrShun(handcards[jiangNumber], jiangNumber == 3)){success = true;break;}else{handcards[jiangNumber][i] += 2;handcards[jiangNumber][0] += 2;}}}return success;}//判斷是否滿足牌組為順子或砍組成public static bool IsKanOrShun(int[] arr, bool isZi){if (arr[0] == 0){return true;}int index = -1;for (int i = 1; i < arr.Length; i++){if (arr[i] > 0){index = i;break;}}bool result;//是否滿足全是砍if (arr[index] >= 3){arr[index] -= 3;arr[0] -= 3;result = IsKanOrShun(arr, isZi);arr[index] += 3;arr[0] += 3;return result;}//是否滿足為順子if (!isZi){if (index < 8 && arr[index + 1] >= 1 && arr[index + 2] >= 1){arr[index] -= 1;arr[index + 1] -= 1;arr[index + 2] -= 1;arr[0] -= 3;result = IsKanOrShun(arr, isZi);arr[index] += 1;arr[index + 1] += 1;arr[index + 2] += 1;arr[0] += 3;return result;}}return false;}}

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