Counting Letters using Arrays

[JavaNooby]

This is Homework so writing out the answer would be unacceptable.

Anyways, The objective is to write a method that will pass a set of test in which it counts the Letters in the test. I have little to no idea what to do and what I got is from messy notes. My professor says that we can us a "helper method" called isLetter. However I do not know how I would define this method in which it would "check" to see if the statements in the test have a certain letter. The reason the size of the array is 27 is because not only does it include the alphabet, but it also include a symbol. Thanks ahead of time.

public int[] letterCounts(String s) {
int[] counts = new int[27];

for (int i= 0; i < s.length(); ){

if(isLetter(s.charAt(i))){
counts[i - 'a']++;

}
else{
}
counts[26]++;
}
return counts;
}
private boolean isLetter(char c){
return true;
}
}

[bgroves85]

If you cast a character as an integer, that character's ascii value will be returned. Uppercase letters have ascii values between 65 and 90 and lowercase letters have ascii values between 97 and 122.

A more complete ascii table can be found here: http://www.asciitable.com/

You may be able to use this in your helper isLetter() method.

Also there are a couple things about the code you posted:

for (int i= 0; i < s.length(); ){

   if(isLetter(s.charAt(i))){
      counts[i - 'a']++;  /* the 'i' here refers to an index into the String, not the character itself */
   }
   else{
   } /* your braces seem to mess up starting here */
   counts[26]++;
}
return counts;
}

[JavaNooby]

If you cast a character as an integer, that character's ascii value will be returned. Uppercase letters have ascii values between 65 and 90 and lowercase letters have ascii values between 97 and 122.

A more complete ascii table can be found here: http://www.asciitable.com/

You may be able to use this in your helper isLetter() method.

Also there are a couple things about the code you posted:

for (int i= 0; i < s.length(); ){

   if(isLetter(s.charAt(i))){
      counts[i - 'a']++;  /* the 'i' here refers to an index into the String, not the character itself */
   }
   else{
   } /* your braces seem to mess up starting here */
   counts[26]++;
}
return counts;
}

So would I switch the i to be a char value?

[bgroves85]

Ok so your integer array called counts is going to count how many of each character there are. Every time you see an 'a' you want to increment the value stored in count[0], when you see a 'b' you increment counts[1], and so on.

String str = "hello";
counts[str.charAt(1) - 'a'] ++;

In that code example, str.charAt(1) will return 'e' because there is an 'e' at index 1 of "hello".

When you use the '-' operator with characters, the characters are converted to their ASCII values. The ASCII value for 'e' is 101 and the ASCII value for 'a' is 97. 101 - 97 = 4. So by doing this you have found the index you need into your counts array.

You had the right idea, but where you used just the 'i' you should have been using the ith index into the string instead.

The reason you would want to make a helper method to determine if a give character is a letter or not is that if you use this approach and the character isn't a letter, you are going to end up with negative indexes into your counts array or indexes greater than 26.

Also you might want to figure out whether a given letter is uppercase or lowercase, because subtracting 'a' from an uppercase letter will not give you what you are looking for.

[dlorde]

Be aware that Java uses Unicode, of which ASCII is an incidental subset. In general, you should not rely on the letters having sequential numeric values - it may be true for US English, but is not necessarily true for other languages. Basically, it is not Good Practice to do math on characters assuming they happen to be in the ASCII range.

If you use the java.lang.Character class, you'll see it provides an isLetter() method that you can pass a char to:

...
if (Character.isLetter(str.charAt(i))) {
 ...
}

YMMV.

The key to performance is elegance, not batallions of special cases...
J. Bently & D. McIlroy