CODEWITHSUNDEEP
java
BodyBingers
BodyBingers

Reputation: 65

How can I use several conditions in a loop to compare different indexes of a string?

so I have this mini-project I'm working for myself on and basically it's a simple method in java for a card game that checks if the string has a flush. Keep in mind I'm a beginner in java(and cardgames/poker...) so I"m new to all of this.

e.g: the string is ""Kh3h7h8h2s" so the hand is King of hearts, 3 of hearts, 7 of hearts, 8 of hearts, 2 of spades etc. The rank is in the 1st, 3rd, 5th, 7th, 9th index of the string. The string needs to have 5 of the same rank to have a flush 

public class CardProblemss {

String hand = "Kh3h7h8h2h";

public boolean hasFlush(String hand) {
    String suit1 = Character.toString(hand.charAt(1));
    String suit2 = Character.toString(hand.charAt(3));
    String suit3 = Character.toString(hand.charAt(5));
    String suit4 = Character.toString(hand.charAt(7));
    String suit5 = Character.toString(hand.charAt(9));
    int flushcounter = 0;

    while (true) {
    if (suit1.equals(suit2)) {
        flushcounter++;
    }            
    else if (suit2.equals(suit3)) {
        flushcounter++;
    } 
    else if (suit3.equals(suit4)) {
        flushcounter++;
    } 
    else if (suit4.equals(suit5)) {
        flushcounter++;
    } 
    System.out.println(flushcounter);
     if (flushcounter >= 5) {
        return true;
    }
    else {
        return false;
    }
    }  

}

Now I can already tell my code is bleh. But my question is, can I formulate a loop that would basically use the charAt(index) method by itself? Since the rank of the hand is at index 1,3,5,7,9, can I write a loop that would increase the index of this method by 2 every time it's ran? Or is there a more efficient way I can do this? Thanks alot.

Upvotes: 0

Views: 119

Answers (2)

Nefari0uss
Nefari0uss

Reputation: 41

Would something like this work?

int flushCounter = 0;

char s = hand.charAt(1); // s for suite
for (int i = 3; i < hand.length(); i = i + 2) {
  if (s == hand.charAt(i)) {
    flushCounter++;
  }
}

if (flushCounter >= 5) {
  return true;
} else {
  return false;
}

This way, the only way you'll iterate through the string and then on ever odd value, you'll compare the suite. I'm assuming that you're hand size will always be 3 or more. If you want i to start at 1, you can do that too. 

int flushCounter = 0;
char s; // s for suite

for (int i = 1; i < hand.length(); i = i + 2) {
  if (i == 1) {
    s = hand.charAt(1); 
  } else if (s == hand.charAt(i)) {
    flushCounter++;
  }
}

if (flushCounter >= 5) {
  return true;
} else {
  return false;
}

With the latter you have to be careful not to initialize s inside the if statement otherwise it'll go out of scope before you can use it.

Note:

if (flushCounter >= 5) {
  return true;
} else {
  return false;
}

can be reduced to

return flushCounter >= 5;

as it will evaluate the expression and then return the result. The former way is much more verbose. Since you are simply returning the value of the expression, I would argue that the latter is simpler.

Upvotes: 1

flakes
flakes

Reputation: 23654

Just for fun, here's how you can do this with streams! 

public static boolean hasFlush(final String hand) {
    final char first = hand.charAt(1);
    final long count = hand.chars().filter(c -> c == first).count();
    return count == 5;
}

Upvotes: 0

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