Java program to count occurence of letters

TheRiddler

Hi this is my first time posting in this forum so forgive any mistakes! Okay so I'm learning Java in college and I need help with a program. The idea for the program is for the user to input a sentence, and the program will count how many times each letter is in the string, ie. "Abba" will give the results:
A=2
B=2

I've made the input string into an array and my method is to search the array for letters. Here is the code so far:



public class SentenceProgram {
public static void main (String args[]) {

Terminal terminal;
String Sentence;
char [] characterArray;

terminal = new Terminal("Sentence Program");
String stringSentence = terminal.readString("Enter in sentence: ");
terminal.println("");


char[] chars = stringSentence.toCharArray();

/*count how many times 'A' is in the sentence*/
int countA = 0;
for(int i = 0; i < chars.length; i++) {
if(chars == 'A')
countA++;
if (chars == 'a')
countA++;
}
if (countA != 0)
terminal.println ("Letter 'A' occurs: " + countA + " times");


}
}



My problem is I'm not sure how to make a method to check every letter. In my code I have a method to check for upper and lower case a but I don't want to have to write this 26 times for every letter. Does anyone know how to write a method to search the array for every letter? Thanks.

kenbrady

string = string.trim();
removes trailing leading white spaces, improves performance.

string = string.toLowerCase();
Now you don't have to check A and a

Have an array of the alphabet and loop through it to test.
letters[]

Have an array of ints to represent the occurances. Set all to 0 first.
occur[]


Then the statement to print the results is
String space = " = ";
for(int i=0;i<26;i++){
terminal.println(letters+space+occur);
}

TheRiddler

Great got it working thanks! Mods you can close this now

Evil Phil

Nah we'll leave it open as somebody else may have input or a question. I'll change it to [Solved] though.

GreeBo

You could also do something like

Stringp[] alphas = new String {"a","b",...."z"};
myString = "abcdshkjbdsd";

len = myString.length();

for (string char : alphas)
myString = myString.replace(char, "");
newLen = myString.length();
print("num "+char+": "+len - newLen);
len = newLen;
done

One benefit is that the string you are checking gets shorter and shorter each time.

km991148

Or sort it alphabetically (after doing tolower on it) then work through it as follows:

for curr=0; curr<totallength; curr++{
{
while(curr!= curr+1)
{count it}
display letter and count
update curr
}


Sorry I was bored - one of these ones with a hundred ways of implementing..
Not even sure if there is any benefit to the above due to the sort..

GreeBo

km991148 wrote: »

Sorry I was bored - one of these ones with a hundred ways of implementing..
Not even sure if there is any benefit to the above due to the sort..

Is there a better reason to be on here?

I think the sort would kill you there though...

km991148

GreeBo wrote: »

Is there a better reason to be on here?

I think the sort would kill you there though...

ha,

ye to both..

But then Sorting is a whole other exercise for the OP..

ronivek

An ideal procrastination opportunity.

public static void printAlphabetCharOccurrences(String str)
{
    int modifier = 97;
    int[] charCount = new int[('z' - 'a') + 1];
    String lowerCaseStr = str.toLowerCase();
    char c;
    for (int i = 0; i < lowerCaseStr.length(); i++) {
        if (Character.isLetter(c = lowerCaseStr.charAt(i))) {
            charCount[c - modifier]++;
        }
    }

    System.out.println("Occurrences of each letter in string: " + str);
    for (int i = 0; i < charCount.length; i++)
        System.out.println((char)(i + modifier) + ": " + charCount[i]);
}