I need to count the number of words and I am assuming the correct way to do it is by calculating the number of times that the previous character in a string is not a letter (ie other characters) because this is to assume that there would be colons,spaces,tabs, and other signs in the string. So at first my idea was to loop through each character and count how many times that you will not get a letter of an alphabet
for(int i = 0; i < string.length(); i++) {
for(int j = 0; i < alphabets.length(); j++) {
if (string.charAt(i-1) == alphabets.charAt(j)) {
counter++;
}
}
}
However I will always get an array out of bounds because of this. So, I kinda need a little help or another way that can actually be more efficient. I thought of using Matches to only [a-zA-z] but I'm not sure how do I handle a char to be comparable to a string in counting how many times it occurs.
Thank you
Addressing the code directly, your first loop has i=0 as the first value of i, but then you ask for
which is where your array-out-of-bounds is coming from.
The second loop has another problem:
You may also want to consider apostrophes as parts of words as well.
The reason you are getting an
IndexOutOfBoundsException
is probably because when i is 0 your inner loop will havestring.charAt(i-1)
which will throw an exception since 0-1 is -1. If you fix that your method might work, although you can use more efficient techniques.You are incrementing the counter if the character is some alphabet character. You should increment it if it is no alphabet character.
Your suggestion to use a regex like "[A-Za-z]" would work fine. In a split command, you'd split on the inverse, like:
String[] words = "Example test: one, two, three".split("[^A-Za-z]+");
EDIT: If you're just looking for raw speed, this'll do the job more quickly.
Use just like this
You can use String.split() to convert the string into an array, with one word in each element. The number of words is given by the length of the array: