十三水中各种牌型判断LUA版
近期回归程序行业，由于业务需求需要做十三水游戏，什么是十三水就不在多讲，下面是判断十三水牌型的方法（带大小王）
GetSSSPaiType = {};
local this = GetSSSPaiType;
local huaseTable = {};
local numTable = {};
function GetSSSPaiType.Gtetype(cardList)
--首先清空数组
huaseTable = {};
numTable = {};
--格式化牌
for i,card in ipairs(cardList) do
local num,huase = GetPokerHasJoker(card)
table.insert(huaseTable,huase);
table.insert(numTable,num);
end
--获取是第几墩牌需要判断
if(#cardList==3)then --是头道 只判断是三条顺子或者散牌 头道只需要判断是三条还是对子
return SSSPaiData[this.GtetSanZhangype(numTable)].imageName;
elseif (#cardList==5)then
return SSSPaiData[this.GtetWuZhangType(huaseTable,numTable)].imageName;
else
return SSSPaiData[SSSPaiType.Error].imageName; --数组越界 return
end
end
function GetSSSPaiType.GtetSanZhangype(cards) --判断头道是否三条
if cards[1] == cards[2] and cards[1]==cards[3]then --是三条
return SSSPaiType.SanTiao;
else
if cards[1] == cards[2] or cards[1] == cards[3] or cards[3] == cards[2] then --是对子
return SSSPaiType.DuiZi;
else --是散牌
return SSSPaiType.SanPai;
end
end
end
--牌大小优先级 五同>同花顺>炸弹>葫芦>同花>顺子>三条>两对>对子>散牌
function GetSSSPaiType.GtetWuZhangType(huaNums,dianNums) --五张牌需要两个参数 花色和点数
if(table.ContainsValue(dianNums,0))then --带王 特殊判断
table.RemoveValue(dianNums, 0) --首先吧王移除
table.RemoveValue(huaNums, 5) --花色也需要移除
if (table.ContainsValue(dianNums,0))then --再次判断是否有王 防止有两个王 --有两个王
table.RemoveValue(dianNums, 0) --有两个王再次移除掉王
table.RemoveValue(huaNums, 5) --花色也需要移除
if(this.isWuTongHasTwoJoker(dianNums))then --两个王 是五同
return SSSPaiType.WuTong;
elseif(this.isTongHuaHasTwoJoker(huaNums) and this.isShunZi(dianNums) and table.getrepeat(dianNums)==3)then --同花顺
return SSSPaiType.TongHuaShun;
elseif(this.isTeShuShunZiHasTwoJoker(dianNums) and this.isTongHuaHasTwoJoker(dianNums)) then --特殊同花顺
return SSSPaiType.ShunZi;
elseif(this.isZhaDanHasTwoJoker(dianNums))then --是炸弹
return SSSPaiType.ZhaDan;
elseif(this.isTongHuaHasTwoJoker(dianNums))then --同花
return SSSPaiType.TongHua;
elseif(this.isShunZi(dianNums))then --顺子
return SSSPaiType.ShunZi;
elseif(this.isTeShuShunZiHasTwoJoker(dianNums))then --特殊顺子
return SSSPaiType.ShunZi;
else
return SSSPaiType.SanTiao; --有两个王 最次也是个三条 不能是葫芦
end
else --只有一个王
if(this.isWuTongHasOneJoker(dianNums))then --一个王 是五同
return SSSPaiType.WuTong;
elseif (this.isTongHuaHasOneJoker(huaNums) and this.isShunZi(dianNums) and table.getrepeat(dianNums)==4)then --是同花顺
return SSSPaiType.TongHuaShun;
elseif(this.isTeShuShunZiHasOneJoker(dianNums)and this.isTongHuaHasOneJoker(huaNums))then --特殊同花顺
return SSSPaiType.TongHuaShun;
elseif(this.isZhaDanHasOneJoker(dianNums))then --是炸弹
return SSSPaiType.ZhaDan;
elseif(this.isHuLuHasOneJoker(dianNums))then --是葫芦
return SSSPaiType.HuLu;
elseif(this.isTongHuaHasOneJoker(huaNums) and not this.isShunZi(dianNums))then --是同花 但不是顺子
return SSSPaiType.TongHua;
elseif(this.isShunZi(dianNums) and not this.isTongHuaHasOneJoker(huaNums) and table.getrepeat(dianNums)==4)then --是顺子 但不同花
return SSSPaiType.ShunZi;
elseif(this.isTeShuShunZiHasOneJoker(dianNums)and not this.isTongHuaHasOneJoker(huaNums))then --特殊顺子 但不是同花
return SSSPaiType.ShunZi;
elseif(this.isSnaTiaoHasOneJoker(dianNums))then --是三条
return SSSPaiType.SanTiao;
elseif(this.isLiangDuiHasOneJoker(dianNums))then --是两对
return SSSPaiType.LiangDui;
else
return SSSPaiType.DuiZi; --有一个王 最低也是个对子
end
end
else --不带王 直接判断
if(this.isWuTong(dianNums))then --是五同
return SSSPaiType.WuTong;
elseif(this.isShunZi(dianNums) and this.isTongHua(huaNums))then --是同花顺
return SSSPaiType.TongHuaShun;
elseif(this.isTeShuShunZi(dianNums) and this.isTongHua(huaNums))then --特殊同花顺
return SSSPaiType.ShunZi;
elseif(this.isZhaDan(dianNums))then --炸弹
return SSSPaiType.ZhaDan;
elseif(this.isHuLu(dianNums)) then --是葫芦
return SSSPaiType.HuLu;
elseif(this.isTongHua(huaNums))then --是同花
return SSSPaiType.TongHua;
elseif(this.isShunZi(dianNums) and table.getrepeat(dianNums)==5)then --只有连续的数才是顺子 不能有重复的
return SSSPaiType.ShunZi;
elseif(this.isTeShuShunZi(dianNums))then --12345也是顺子
return SSSPaiType.ShunZi;
elseif(this.isSanTiao(dianNums))then --是三条
return SSSPaiType.SanTiao;
elseif(this.isLiangDui(dianNums))then --是两对
return SSSPaiType.LiangDui;
elseif(this.isDuiZi(dianNums))then --不对子
return SSSPaiType.DuiZi;
else
return SSSPaiType.SanPai;
end
end
end
--判断是否顺子 带王
function GetSSSPaiType.isShunZi(paiNums)
local n = #paiNums
--计算0的数量
local joker = 0
for i=1,n do
if paiNums[i] == 0 then
joker = joker + 1
end
end
--排序
local sortt = function(a, b)
return a < b
end
table.sort(paiNums,sortt)
local pre = paiNums[1]
--计算总的补充数量
local x = 0
for i=2,n do
local tx = paiNums[i] - pre
x = x + tx - 1
pre = paiNums[i]
end
--如果x比0的数量少，那么可以将数列补充成连续的
if x <= joker then --是顺子
return true;
else
--不是顺子
return false;
end
end
--12345也是顺子 特殊判断
function GetSSSPaiType.isTeShuShunZi(paiNums)
local teshu = {2,3,4,5,14};
table.sort(paiNums)
if(table.isEquation(teshu,paiNums))then
return true;
else
return false;
end
end
--是否同花
function GetSSSPaiType.isTongHua(huaNums)
if(huaNums[1]==huaNums[2] and huaNums[1]==huaNums[3] and huaNums[1]==huaNums[3] and huaNums[1]==huaNums[4] and huaNums[1]==huaNums[5]) then
return true;
else
return false;
end
end
--是否五同
function GetSSSPaiType.isWuTong(paiNums)
if(paiNums[1]==paiNums[2] and paiNums[1]==paiNums[3] and paiNums[1]==paiNums[3] and paiNums[1]==paiNums[4] and paiNums[1]==paiNums[5]) then
return true;
else
return false;
end
return false;
end
--判断是否葫芦
function GetSSSPaiType.isHuLu(paiNums)
if paiNums[1] == paiNums[2] and paiNums[1] == paiNums[3] then
if paiNums[4] == paiNums[5] then
return true
end
end
if paiNums[3] == paiNums[4] and paiNums[3] == paiNums[5] then
if paiNums[1] == paiNums[2] then
return true
end
end
return false
end
--判断是否三条
function GetSSSPaiType.isSanTiao(paiNums)
table.sort(paiNums)
if(paiNums[1]==paiNums[2]and paiNums[2]==paiNums[3])then
return true;
elseif(paiNums[2]==paiNums[3]and paiNums[3]==paiNums[4])then
return true;
elseif(paiNums[3]==paiNums[4]and paiNums[4]==paiNums[5])then
return true;
else
return false
end
end
--判断是否炸弹
function GetSSSPaiType.isZhaDan(paiNums)
table.sort(paiNums) --避免浪费性能 排序后单牌只会在1或者5
if paiNums[1] == paiNums[2] and paiNums[1] == paiNums[3] and paiNums[1] == paiNums[4] then
return true
elseif paiNums[2] == paiNums[3] and paiNums[2] == paiNums[4] and paiNums[2] == paiNums[5] then
return true
else
return false
end
end
--判断是否两对
function GetSSSPaiType.isLiangDui(paiNums)
table.sort(paiNums) --避免浪费性能 排序后单牌只会在1 3 5
if paiNums[1] == paiNums[2] and paiNums[4] == paiNums[5] then
return true
elseif paiNums[2] == paiNums[3] and paiNums[4] == paiNums[5] then
return true
elseif paiNums[1] == paiNums[2] and paiNums[2] == paiNums[3] then
return true
else
return false
end
end
--判断是否是对子 没有王
function GetSSSPaiType.isDuiZi(paiNums)
table.sort(paiNums)
if(paiNums[1]==paiNums[2] or paiNums[2]==paiNums[3] or paiNums[3]==paiNums[4] or paiNums[4]==paiNums[5])then
return true
else
return false
end
end
-------------------------有一个王判断-------------------------
--判断是否五同 有一个王
function GetSSSPaiType.isWuTongHasOneJoker(paiNums)
if (paiNums[1]==paiNums[2] and paiNums[1]==paiNums[3] and paiNums[1]==paiNums[4])then
return true
else
return false
end
end
--判断是否同花 有一个王
function GetSSSPaiType.isTongHuaHasOneJoker(huaNums)
if(huaNums[1]==huaNums[2] and huaNums[1]==huaNums[3] and huaNums[1]==huaNums[4])then
return true
else
return false
end
end
--判断是否特殊顺子 有一个王 就是和王能组成A2345的顺子
function GetSSSPaiType.isTeShuShunZiHasOneJoker(paiNums)
table.sort(paiNums) --首先吧自己的手牌格式化
local teshu1 = {2,3,4,5};
local teshu2 = {3,4,5,14};
local teshu3 = {2,4,5,14};
local teshu4 = {2,3,5,14};
local teshu5 = {2,3,4,14};
if(table.isEquation(teshu1,paiNums) or table.isEquation(teshu2,paiNums) or table.isEquation(teshu3,paiNums) or table.isEquation(teshu4,paiNums) or table.isEquation(teshu5,paiNums))then
return true;
else
return false;
end
end
--判断是否是炸弹 有一个王
function GetSSSPaiType.isZhaDanHasOneJoker(paiNums)
table.sort(paiNums) --格式化牌 让散牌在1或者4
if(paiNums[1]==paiNums[2] and paiNums[1]==paiNums[3])then
return true;
elseif(paiNums[2]==paiNums[3] and paiNums[2]==paiNums[4])then
return true;
else
return false;
end
end
--判断是否是葫芦 有一个王
function GetSSSPaiType.isHuLuHasOneJoker(paiNums)
table.sort(paiNums)
if(paiNums[1]==paiNums[2] and paiNums[3] == paiNums[4])then
return true;
else
return false;
end
end
--判断是否三条 有一个王
function GetSSSPaiType.isSnaTiaoHasOneJoker(paiNums)
table.sort(paiNums)
if paiNums[1] == paiNums[2] or paiNums[2] == paiNums[3] or paiNums[3]==paiNums[4] then
return true
else
return false
end
end
--判断是否两对 有一个王
function GetSSSPaiType.isLiangDuiHasOneJoker(paiNums)
table.sort(paiNums)
if(paiNums[1]==paiNums[2] or paiNums[2]==paiNums[3] or paiNums[3]==paiNums[4])then
return true
else
return false
end
end
-------------------------有两个王判断-------------------------
--判断是否五同
function GetSSSPaiType.isWuTongHasTwoJoker(paiNums)
if(paiNums[1] == paiNums[2] and paiNums[1] == paiNums[3])then
return true;
else
return false;
end
end
--判断是否同花
function GetSSSPaiType.isTongHuaHasTwoJoker(huaNums)
if(huaNums[1] == huaNums[2] and huaNums[1] == huaNums[3])then
return true;
else
return false;
end
end
--判断是否炸弹
function GetSSSPaiType.isZhaDanHasTwoJoker(paiNums)
table.sort(paiNums)
if(paiNums[1] == paiNums[2] or paiNums[2] == paiNums[3])then
return true;
else
return false;
end
end
--特殊顺子 就是和王能组成A2345的顺子
function GetSSSPaiType.isTeShuShunZiHasTwoJoker(paiNums)
table.sort(paiNums)
local teshu1 = {2,3,14};
local teshu2 = {2,4,14};
local teshu3 = {2,5,14};
local teshu4 = {3,4,14};
local teshu5 = {3,5,14};
local teshu6 = {2,3,4};
local teshu7 = {2,3,5};
local teshu8 = {2,4,5};
local teshu9 = {3,4,5};
if(table.isEquation(teshu1,paiNums) or table.isEquation(teshu2,paiNums) or table.isEquation(teshu3,paiNums) or table.isEquation(teshu4,paiNums) or table.isEquation(teshu5,paiNums) or table.isEquation(teshu6,paiNums) or table.isEquation(teshu7,paiNums) or table.isEquation(teshu8,paiNums) or table.isEquation(teshu9,paiNums))then
return true;
else
return false;
end
end
使用时仅需调用GetSSSPaiType.Gtetype()即可，牌类型如下：
用到的牌型枚举：
SSSPaiData = {};
SSSPaiData[SSSPaiType.SanPai] = {typeName = "散牌",imageName = "sanpai"};
SSSPaiData[SSSPaiType.DuiZi] = {typeName = "对子",imageName = "duizi"};
SSSPaiData[SSSPaiType.LiangDui] = {typeName = "两对",imageName = "liangdui"};
SSSPaiData[SSSPaiType.SanTiao] = {typeName = "三条",imageName = "santiao"};
SSSPaiData[SSSPaiType.ShunZi] = {typeName = "顺子",imageName = "shunzi"};
SSSPaiData[SSSPaiType.TongHua] = {typeName = "同花",imageName = "tonghua"};
SSSPaiData[SSSPaiType.HuLu] = {typeName = "葫芦",imageName = "hulu"};
SSSPaiData[SSSPaiType.ZhaDan] = {typeName = "炸弹",imageName = "zhadan"};
SSSPaiData[SSSPaiType.TongHuaShun] = {typeName = "同花顺",imageName = "tonghuashun"};
SSSPaiData[SSSPaiType.WuTong] = {typeName = "五同",imageName = "tiezhi"};
SSSPaiData[SSSPaiType.Error] = {typeName = "错误",imageName = "error"};用到的公共方法：可以放本类，也可以放全局，个人认为放全局会方便一点，这样其他地方也可以使用全局进行调用
--切割牌 带王
function GetPokerHasJoker(pkId)
local sets = string.split(pkId, "_");
local color = tonumber(sets[1]);
local num = tonumber(sets[2]);
if(num == 1) then
num = 14;
end
if(color == 5)then --如果是王 就吧点数当成0
num = 0;
end
return num, color;
end
--判断value是否存在
--字典/数组
function table.ContainsValue(tb, val)
if(type(tb) == "table") then
for k, v in pairs(tb) do
if equals(v, val) then
return true;
end
end
else
return equals(tb, val);
end
return false;
end
--删除字典中值
function table.RemoveValue(tb, val)
for k, v in pairs(tb) do
if(v == val) then
table.remove(tb, k);
return;
end
end
end
--判断两个数组是否相等
function table.isEquation(tb1,tb2)
if(#tb1~=#tb2)then
return false;
end
for i,v in pairs(tb1) do
if tb1[i] ~= tb2[i]then
return false;
end
end
return true;
end
--检查表内相同元素个数
function table.getrepeat(t)
local check = {}
local n = {}
for key, value in pairs(t) do
if not check[value] then
n[key] = value
check[value] = value
end
end
return #n;
end写在后面：方法是临时拼凑，可能会有bug，我会持续修复，直至修复完美！！