Function to determine whether a poker hand is a straight?

Question

For a homework assignment I was given a Card class that has enumerated types for the Rank and Suit. I am required to compare two poker hands (each hand is an ArrayList of 5 cards) and decide the winner.

The isStraight() function is really bothering me, because I have to start over the count after the Ace. For example,

QUEEN, KING, ACE, TWO, THREE

Is still considered a straight. What is the best way to code this functionality?

Here is the Rank/Suit enumerated type code, if that helps.

public enum Rank
{
    TWO(2), THREE(3), FOUR(4), FIVE(5), SIX(6), SEVEN(7), EIGHT(8), NINE(9),
    TEN(10), JACK(11), QUEEN(12), KING(13), ACE(14);

    private final int points;

    private Rank(int points)
    {
        this.points = points;
    }

    public int points()
    {
        return this.points;
    }
}

public enum Suit
{
    DIAMONDS, CLUBS, HEARTS, SPADES;
}

Answer 1

You do realize that by the rules of any poker game I've ever played or heard of a straight cannot wrap right? Ace can be low [A,2,3,4,5] or high [10,J,Q,K,A] but it can't wrap. According to those rules (not yours) I've implemented something similar before. Basically you sort the array and walk it, making sure the current card is one higher than the previous. In the first iteration, if it is an ace, then you explicitly check for [A,2,3,4,5]. If it is you return true and if it isn't you continue with the normal straight logic. This should set you in the right direction.