确定扑克 Table 个位置 - Sit'n Go 锦标赛
Determine Poker Table Positions - Sit'n Go Tournaments
Note: I do the javascript code according to the ajrwhite answer. Hope it helps someone.
Link: http://codepen.io/eMineiro/pen/EKrNBe
Open codepen console to see the examples working.
在扑克中，我们根据庄家来定义玩家位置。像这样：
蓝色：小盲注和大盲注位置
绿色：迟到 Dealer/Late 个位置
黄色：中间位置
粉红色：早期位置
所以，假设这两个数组：
players:[1,2,3,4,5,6,7,8,9,10];
positions:["bb","sb","btn","late","medium","medium","medium","early","early","early"];
在这种情况下，"player1"就是"Big Blind"，"player2"就是"Small Blind"，"player3"就是"button"... ..
我想在调用 changePositions(dealer) 时对玩家数组进行排序。示例：
changePosition(10); //means that "player10" now is the new Dealer
结果应该是：
players:[2,1,10,9,8,7,6,5,4,3];
positions:["bb","sb","btn","late","medium","medium","medium","early","early","early"];
游戏过程中，玩家可能会被淘汰。所以我有一个功能可以排除 "positions array" 中的 "last position" 并排除播放器。然后我需要再次调用 changePosition(X)，其中 X 是 "player10"（实际庄家）左侧的下一个未被淘汰的玩家。
已删除 "player 1" 的示例，新数组应为：
players:[2,10,9,8,7,6,5,4,3];
positions:["bb","sb","btn","late","medium","medium","medium","early","early"];
我需要再次调用 changePosition(X) 来确定新的仓位，在本例中为 X=2，因为 "player2" 在实际经销商 "player10"[=23 的左边=]
changePosition(2);
结果应该是：
players:[4,3,2,10,9,8,7,6,5];
positions:["bb","sb","btn","late","medium","medium","medium","early","early"];
玩家淘汰后如何找到新的庄家？
Note: I created a function named changeNextDealer(). Negative index was not the problem, because the next dealer is clockwise. It's in the code pen link.
dealerArrayPosition-1; //But if bigBlind and smallBlind was eliminated simultaneously I get a negative position.
如何将负索引（如 -1）映射到最后一个位置。或者 -2 到 LastPosition-1？有什么快速的方法吗？
Note: This question is still no answered, but is not the main question of this discussion. I think a will ask in a separated post.
changePosition(dealer)函数应该怎么做？
我已经尝试了很多，但不知道该怎么做。
Note: I created a function named changePosition(). It's in the codepen link.
让我们假设球员总是编号为 1-10。我们不需要两个数据结构来存储玩家的位置和状态。我们只需要一个索引数据结构来存储玩家状态：
状态 1
positions:["btn","sb","bb","early","early","early","medium","medium","medium","late"];
状态 2
positions: ["late","btn","sb","bb","early","early","early","medium","medium","medium"]
状态 3
positions: ["eliminated","late","btn","sb","bb","early","early","early","medium","medium"];
这存储的信息量与您的两个数组相同，并且更加一致。您只需访问 positions[0] 查看玩家 1 的状态或 positions[9] 查看玩家 10.
现在有了这个更一致的结构，应该更容易处理每一轮结束时的状态变化。
淘汰玩家
每一轮结束后，将一名被淘汰的玩家的状态更改为"eliminated"。此时您可能需要一个临时数组，以便您可以检查上一轮发生的事情仍然会为下一轮更新信息。
如何把数组看成一个圆
将庄家筹码 "btn" 移动到 (currentPositionOfBtn+1)%10 - 这意味着按钮从 1 移动到 2 再到 3，等等，但是通过使用余数运算符和玩家总数，我们确保从位置 9（玩家 10）的移动循环回到 0（玩家 1）。
如果 currentPositionOfBtn+1)%10 中的玩家已被淘汰，则检查 currentPositionOfBtn+2)%10 等等 - 这是一个易于实现的循环。
重新计算位置
现在鉴于玩家被淘汰时位置的描述发生变化（到最后你没有 "early" 或 "medium"），我建议从庄家开始重新计算位置每轮结束时，跳过任何标记为 "eliminated".
的玩家
Note: I do the javascript code according to the ajrwhite answer. Hope it helps someone.
Link: http://codepen.io/eMineiro/pen/EKrNBe
Open codepen console to see the examples working.
在扑克中，我们根据庄家来定义玩家位置。像这样：
蓝色：小盲注和大盲注位置
绿色：迟到 Dealer/Late 个位置
黄色：中间位置
粉红色：早期位置
所以，假设这两个数组：
players:[1,2,3,4,5,6,7,8,9,10];
positions:["bb","sb","btn","late","medium","medium","medium","early","early","early"];
在这种情况下，"player1"就是"Big Blind"，"player2"就是"Small Blind"，"player3"就是"button"... ..
我想在调用 changePositions(dealer) 时对玩家数组进行排序。示例：
changePosition(10); //means that "player10" now is the new Dealer
结果应该是：
players:[2,1,10,9,8,7,6,5,4,3];
positions:["bb","sb","btn","late","medium","medium","medium","early","early","early"];
游戏过程中，玩家可能会被淘汰。所以我有一个功能可以排除 "positions array" 中的 "last position" 并排除播放器。然后我需要再次调用 changePosition(X)，其中 X 是 "player10"（实际庄家）左侧的下一个未被淘汰的玩家。
已删除 "player 1" 的示例，新数组应为：
players:[2,10,9,8,7,6,5,4,3];
positions:["bb","sb","btn","late","medium","medium","medium","early","early"];
我需要再次调用 changePosition(X) 来确定新的仓位，在本例中为 X=2，因为 "player2" 在实际经销商 "player10"[=23 的左边=]
changePosition(2);
结果应该是：
players:[4,3,2,10,9,8,7,6,5];
positions:["bb","sb","btn","late","medium","medium","medium","early","early"];
玩家淘汰后如何找到新的庄家？
Note: I created a function named changeNextDealer(). Negative index was not the problem, because the next dealer is clockwise. It's in the code pen link.
dealerArrayPosition-1; //But if bigBlind and smallBlind was eliminated simultaneously I get a negative position.
如何将负索引（如 -1）映射到最后一个位置。或者 -2 到 LastPosition-1？有什么快速的方法吗？
Note: This question is still no answered, but is not the main question of this discussion. I think a will ask in a separated post.
changePosition(dealer)函数应该怎么做？
我已经尝试了很多，但不知道该怎么做。
Note: I created a function named changePosition(). It's in the codepen link.
让我们假设球员总是编号为 1-10。我们不需要两个数据结构来存储玩家的位置和状态。我们只需要一个索引数据结构来存储玩家状态：
状态 1
positions:["btn","sb","bb","early","early","early","medium","medium","medium","late"];
状态 2
positions: ["late","btn","sb","bb","early","early","early","medium","medium","medium"]
状态 3
positions: ["eliminated","late","btn","sb","bb","early","early","early","medium","medium"];
这存储的信息量与您的两个数组相同，并且更加一致。您只需访问 positions[0] 查看玩家 1 的状态或 positions[9] 查看玩家 10.
现在有了这个更一致的结构，应该更容易处理每一轮结束时的状态变化。
淘汰玩家
每一轮结束后，将一名被淘汰的玩家的状态更改为"eliminated"。此时您可能需要一个临时数组，以便您可以检查上一轮发生的事情仍然会为下一轮更新信息。
如何把数组看成一个圆
将庄家筹码 "btn" 移动到 (currentPositionOfBtn+1)%10 - 这意味着按钮从 1 移动到 2 再到 3，等等，但是通过使用余数运算符和玩家总数，我们确保从位置 9（玩家 10）的移动循环回到 0（玩家 1）。
如果 currentPositionOfBtn+1)%10 中的玩家已被淘汰，则检查 currentPositionOfBtn+2)%10 等等 - 这是一个易于实现的循环。
重新计算位置
现在鉴于玩家被淘汰时位置的描述发生变化（到最后你没有 "early" 或 "medium"），我建议从庄家开始重新计算位置每轮结束时，跳过任何标记为 "eliminated".
的玩家