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Sure! This generates a number for every hand, so that a better hand gets a higher number. The resulting number will contain 11 hexadecimal digits:
0x100000 bbbbb
^^^^^^ \____ the hand itself
|||||\_ 1 if "one pair"
||||\__ 1 if "two pairs"
|||\___ 1 if "three of a kind"
||\____ 1 if "full house"
|\_____ 1 if "four of a kind"
\______ 1 if "five of a kind"
For example:
AAAAA: 0x100000 bbbbb
AAAA2: 0x010000 bbbb0
22233: 0x001000 00011
The hand itself is 5 hexadecimal digits for every card, 0 for "2" to b for "ace".
This way the higher combination always has a higher number, and hands with the same combination are ordered by the order of the cards in the hand.
Wow, this is exactly what I did, but in C#. That's cool.
public class Hand
{
public string Cards;
public int Rank;
public int Bid;
}
public static HandType IdentifyHandType(string hand)
{
var cardCounts = hand
.Aggregate(
new Dictionary(),
(counts, card) =>
{
counts[card] = counts.TryGetValue(card, out var count) ? (count + 1) : 1;
return counts;
})
.OrderByDescending(kvp => kvp.Value);
using (var cardCount = cardCounts.GetEnumerator())
{
cardCount.MoveNext();
switch (cardCount.Current.Value)
{
case 5: return HandType.FiveOfAKind;
case 4: return HandType.FourOfAKind;
case 3: { cardCount.MoveNext(); return (cardCount.Current.Value == 2) ? HandType.FullHouse : HandType.ThreeOfAKind; }
case 2: { cardCount.MoveNext(); return (cardCount.Current.Value == 2) ? HandType.TwoPairs : HandType.OnePair; }
}
}
return HandType.HighCard;
}
public static Hand SetHandRank(Hand hand, Dictionary cardValues)
{
int rank = 0;
int offset = 0;
var cardValueHand = hand.Cards;
for (int i = cardValueHand.Length - 1; i >= 0; i--)
{
var card = cardValueHand[i];
var cardValue = cardValues[card];
var offsetCardValue = cardValue << offset;
rank |= offsetCardValue;
offset += 4; // To store values up to 13 we need 4 bits.
}
// Put the hand type at the high end because it is the most
// important factor in the rank.
var handType = (int)IdentifyHandType(hand.Cards);
var offsetHandType = handType << offset;
rank |= offsetHandType;
hand.Rank = rank;
return hand;
}
I think you have the same solution, basically, just the details are a bit different. I like how you handled the joker, I didn't realise you could just multiply your best streak of cards to get the best possible combination.
I didn't multiply the streak, I just took the jokers and added them to the highest hand already in the list. Is that not what you did? It looked the same to me.