在不使用库的情况下计算字符串中字母出现的次数

Question

My professor says I can't use libraries and stuff. I have my code with libraries:

String phrase = keyboard.nextLine(); //Input.
int addr = phrase.length();
Map<Character, Integer> numChars = new HashMap<Character, Integer>(Math.min(addr, 26)); //there are 26 character in our alphabet. It makes a map and maps our the characters.

for (int i = 0; i < addr; ++i)//this area reads the string then, tells how many times each character appeared. Loops for each chracter.
{
  char charAt = phrase.charAt(i);

  if (!numChars.containsKey(charAt))
  {
    numChars.put(charAt, 1);
  }
  else if (numChars.containsKey(charAt))
  {
    numChars.put(charAt, 0);
  }
  else
  {
    numChars.put(charAt, numChars.get(charAt) + 1);
  }
}
  System.out.println(phrase);//outputs phrase written by user.
  System.out.println(numChars);//outputs results of the code above

// this code reads which one appeared the most.
 int FreqChar = 0;
 char frequentCh = ' ';

 for (int f = 0; f < phrase.length(); f++)
 {
    char poop = phrase.charAt(f);
    int banana = 0;
    for (int j = phrase.indexOf(poop); j != -1; j = phrase.indexOf(poop, j + 1))
    {
        frequentCh++;
    }
    if (banana > FreqChar)
    {
        FreqChar = banana;*/

Here is my program without the libraries so far. I need help translating this into arrays.

    import java.util.*;

  public class LetCount
  {

  public static final int NUMCHARS = 26; //26 chars in alphabet.

  // int addr(char ch) returns the equivalent integer address for the letter
  // given in ch, 'A' returns 1, 'Z' returns 26 and all other letters return
  // their corresponding position as well. felt this is important.

  public static int addr(char ch)
  {
    return (int) ch - (int) 'A' + 1;
  }
// Required method definitions for (1) analyzing each character in an input
// line to update the appropriate count; (2) determining most frequent letter; 
// (3) determining least frequent letter; and (4) printing final results
// should be defined here.

  public static void main(String[] args)
  {
    Scanner keyboard = new Scanner(System.in);  // for reading input

    int[] count = new int [NUMCHARS]; // count of characters

    String phrase = keyboard.nextLine(); //Input.
    int addr = phrase.length();

       for(char ch = 'A'; ch <= 'Z'; ch++)
        {

    }
  }

Answer 1

This is easier to put here than in comments, but it's not really an answer :)

You've got a great start - rather than going through the letters and finding matches, go through the string and increment your letter counter each time you encounter a letter (yeah, that sounds weird in my head too).

Convert the string to lower or upper case first, you only need to count the letter, not whether it's lower or upper based upon your existing code.

Answer 2

If you just have to count all the alphabets in a given String and store it in an array, try this:

String str = phrase.toUpperCase();    
for(int i = 0; i < str.length(); i++) {
    char c = str.charAt(i);
    int charPositionInArray = c - 'A';
    if(charPositionInArray < 26){
        count[charPositionInArray] += 1;
    }
}

Also, the index of an array starts from 0, so I assume you want the count for 'A' to be stored in count[0], ie, the first position in the array.

Also, this code does nothing for any character that is not an alphabet.