HiddenWord free response problem from the 2015 AP Computer Science A Exam.
HiddenWord is #2 from the from the 2015 AP Computer Science A Free Response problems.
https://secure-media.collegeboard.org/digitalServices/pdf/ap/ap15_frq_computer_science_a.pdf
public class HiddenWord
{
private String word;
public HiddenWord(String word)
{
this.word = word;
}
public String getHint(String guess)
{
String hint = "";
for(int i = 0; i < word.length(); i++)
{
String guessLetter = guess.substring(i, i + 1);
if(word.substring(i, i + 1).equals(guessLetter))
hint += guessLetter;
else if(word.indexOf(guessLetter) != -1)
hint += "+";
else
hint += "*";
}
return hint;
}
}
If I used word.CharAt(i) instead of word.substring(i,i+1) would it still work?
You can use
charAtto obtain a single character but you must properly handle the return value’s type.charAtreturns acharrather than aString.Is it OK if I put else if(word.indexOf(getLetter) == -1) at the end instead of just else
Sure. It’s an unnecessary check but it isn’t wrong.
If I used contains() instead of indexOf(), how many points will be taken off?
contains, if used correctly, would work fine. I don’t have the scoring guide but typically any correct code receives full credit unless it rewrites an accessible method.In getHint() is it ok in the for loop to do for(int i = 0; i < word.length()-1; i++)
Thanks
No, that would skip the last letter.
It could be
for(int i = 0; i <= word.length() – 1; i ++)
Sure, you could use symmetric loop bounds (
<= word.length() - 1) instead of asymmetric bounds (< word.length()).