poker 游戏实现 （python）

今天我们来制作poker 这个牌类游戏

1. understanding（理解问题）

   Start with a vague understanding that you refine into a problem.

2. specify （明确问题）

   Specify how this problem can be made amenable to being coded.

3. design （设计程序）

   Coding

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游戏规则：

所有可能的手牌
        • 0- High Card
        • 1- One Pair
        • 2- Two Pair
        • 3- Three of a Kind
        • 4- Straight
        • 5- Flush
        • 6- Full House
        • 7- Four of a Kind
        • 8- Straight Flush

【注：我们用其对应的数字作为得分】

详细游戏规则自行上维基百科上查看

如何在hands 中存储我们的手牌，手牌由数字和花色组成（可以采用如下方式）：

  sf = ['6C', '7C', '8C', '9C', 'TC'] # Straight Flush
  fk = ['9D', '9H', '9S', '9C' ,'7D'] # Four of a Kind
  fh = ['TD', 'TC', 'TH', '7C', '7D']  #Full House

用何种数据结构来存放这些手牌类型：

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我们最终要判断哪副手牌可以获胜，所以我们需要一个判断胜负的函数

Def poker(hands):  
    "return the best hand"

进一步分析我们要判断大小，那么我们需要对手牌有一个大小的衡量机制

Def hand_rank(hand): 
    "return  the rank of hand"

有了这个大小的度量机制之后

return max(hands, key=hand_rank)   就能返回 best hand

于是我们得到我们的poker 函数

def poker(hands):
    "Return the best hand: poker([hand,...]) => hand"
    return max(hands, key=hand_rank)

接下来我们要完成hand_rank() 函数，因为hand_rank 函数涉及到手牌的类型，所以我们要完成对手牌类型的判断

Straight

straight(ranks):
returns True if the hand is a straight.

Flush

flush(hand):
returns True if the hand is a flush

Kind

kind(n, ranks):
returns the first rank that the hand has exactly n of. For A hand with 4 sevens this function would return 7.

two_pair

two_pair(ranks):
if there is a two pair, this function returns their corresponding ranks as a tuple. For example, a hand with 2 twos and 2 fours would cause this function to return (4, 2).

card_ranks

card_ranks(hand)：
returns an ORDERED tuple of the ranks in a hand (where the order goes from highest to lowest rank).

假定以上函数已经完成，那么我们的hand_rank 函数如下

    def hand_rank(hand):
        ranks = card_ranks(hand)
        if straight(ranks) and flush(hand):            # straight flush
            return (8, max(ranks))
        elif kind(4, ranks):                           # 4 of a kind
            return (7, kind(4, ranks), kind(1, ranks))
        elif kind(3, ranks) and