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Start your free trialArcee Palabrica
8,100 PointsCounting Scrabble Tiles Can't figure out Challenge Task 2 of 2 on
Hi guys anyone who can help me out with this challenge... been stuck for quite a while on Task 2 of 2... this is what I did so far:
public class ScrabblePlayer {
// A String representing all of the tiles that this player has
private String tiles;
public ScrabblePlayer() {
tiles = "";
}
public String getTiles() {
return tiles;
}
public void addTile(char tile) {
tiles += tile;
}
public boolean hasTile(char tile) {
return tiles.indexOf(tile) != -1;
}
public int getCountOfLetter(char letter) {
int a = 0;
for (char x : tiles.toCharArray()) {
if (tiles.indexOf(letter) != -1) {
a += 1;
}
}
return a;
}
}
// This code is here for example purposes only
public class Example {
public static void main(String[] args) {
ScrabblePlayer player1 = new ScrabblePlayer();
player1.addTile('d');
player1.addTile('d');
player1.addTile('p');
player1.addTile('e');
player1.addTile('l');
player1.addTile('u');
ScrabblePlayer player2 = new ScrabblePlayer();
player2.addTile('z');
player2.addTile('z');
player2.addTile('y');
player2.addTile('f');
player2.addTile('u');
player2.addTile('z');
int count = 0;
// This would set count to 1 because player1 has 1 'p' tile in her collection of tiles
count = player1.getCountOfLetter('p');
// This would set count to 2 because player1 has 2 'd'' tiles in her collection of tiles
count = player1.getCountOfLetter('d');
// This would set 0, because there isn't an 'a' tile in player1's tiles
count = player1.getCountOfLetter('a');
// This will return 3 because player2 has 3 'z' tiles in his collection of tiles
count = player2.getCountOfLetter('z');
// This will return 1 because player2 has 1 'f' tiles in his collection of tiles
count = player2.getCountOfLetter('f');
}
}
2 Answers
andren
28,558 PointsI'll give you some hints, if you want the solution outright then just ask me and I'll post it.
The major benefit of a
for each
loop over other loops is that it pulls out each item in the list you give it and makes it available within the loop. In yourfor each
loop you name that variablex
. The first time the loop runsx
will equal the first char in thetiles
String, the second time it will equal the second char, and so on. If you use afor each
loop and don't make use of the variable it creates, then that's a pretty good indication that you are doing something wrong.The solution to this task does not involve using the
indexOf
method. That method just tells you if a character exists within a String, it does not tell you how many unique appearances of that character there are.Most of your code is already correct. You pretty much only need to change the condition of your
if
statement.This is more a note about the task checker than your own code. The task expects you to increase your counter variable (
a
in your case) using the++
operator not+= 1
. So you need to replacea += 1
witha++
.
Petri Laine
Courses Plus Student 858 PointsHint: The char x you created for the loop stores each char in your CharArray one at a time each time it loops through.
Arcee Palabrica
8,100 PointsArcee Palabrica
8,100 Pointsandren Thanks for the clue man... keeping things simple really helps.. thanks for reminding me that... have a great day! thanks thanks again... :) got through the task now...