获取用户输入，放入数组并打印出每个字母的使用次数

Question

I am trying to write this program so that when the user inputs a line of text they are given a chart showing how many times each letter is used. I broke it up into an array but I kept getting an error for "counts[letters[a] == 'a']++;" saying i can't convert a string to a char or a boolean to a int, depending on the way I put it. I can't figure out why it's not all char.

import java.util.*;

public class AnalysisA { //open class

public static String input;
public static String stringA;

public static void main (String args []) { //open main 

    System.out.println("Please enter a line of text for analysis:");

    Scanner sc = new Scanner(System.in);
    input = sc.nextLine();
    input = input.toLowerCase();

    System.out.println("Analysis A:");//Analysis A
    System.out.println(AnalysisA(stringA)); 

} // close main 

public static String AnalysisA (String stringA) { // open analysis A

   stringA = input;

   char[] letters = stringA.toCharArray();

   int[] counts = new int[26];

   for (int a = 0; a < letters.length; a++) { //open for 
     counts[letters[a] == 'a']++;
     System.out.print(counts);
   } //close for 
}

Answer 1

The expression letters[a] == 'a' results in a boolean answer (1 or 0), but you have that indexing an array, which must be an int.

What you're basically telling Java is to do counts[true]++ or counts[false]++ , which makes no sense.

What you really want is a HashMap that maps each character to the amount of times you saw it in the array. I won't put the answer here, but look up HashMaps in Java and you'll find the clues you need.

Answer 2

counts[___] expect an Integer index, whereas your expression letters[a] == 'a' return a boolean

Im guessing you're trying to increment your 'Dictionary' value by 1 each time a letter is met. You can get the index by making letters[a] - 'a'

Because of the order in the ASCII table , letter 'a' which equal 97 if subtracted to another letter, say 'b' which is 98, will produce the index 1, which is the correct position for your base26 'Dictionary'

Extra:

• You should use for (int i = ... for indexing ( i instead of a, its easy to mixed variables up if you name your i ndex like that)
• You must make sure all the characters are lower-case before you start doing this, because as you can see in the table above 'B' - 'a' and 'b' - 'a' are 2 very different things.

Answer 3

If you use a Map you can do this easily without complicating..

Map<Character, Integer> map = new HashMap<Character, Integer>();


import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;
import java.util.Set;

public class AnalysisA { // open class

    public static String input;

    public static void main(String args[]) { // open main

        System.out.println("Please enter a line of text for analysis:");

        Scanner sc = new Scanner(System.in);
        input = sc.nextLine();
        input = input.toLowerCase();
        sc.close();
        System.out.println("Analysis A:");// Analysis A
        System.out.println(Analysis());

    } // close main

    public static String Analysis() { // open analysis A

        Map<Character, Integer> map = new HashMap<Character, Integer>();
        char[] letters = input.toCharArray();
        Integer count;
        for (char letter : letters) {
            count = map.get(letter);
            if (count == null || count == 0) {
                map.put(letter, 1);
            } else {
                map.put(letter, ++count);
            }
        }
        Set<Character> set = map.keySet();
        for (Character letter : set) {
            System.out.println(letter + " " + map.get(letter));
        }
        return "";
    }
}