字符串按降序排列

Question

Does anyone know how to sort a string eg: afacfa into aaaffc? ( descending order of frequency ) I am very new to programming and have only been able to come up with.

String word = (String)jTextField1.getText();
        String indexes = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz";
        int[] count = new int[indexes.length()];
        for (int i = 0; i < word.length(); i++) {
            int index = indexes.indexOf(word.charAt(i));

        System.out.println( "" + i + count[2] ) ;
        if (index < 0)
        continue;

    count[index]++;
        }
        for (int i = 0; i < count.length; i++) {
    if (count[i] < 1)
        continue;

    jTextArea1.append(String.format("%s (%d) %s",indexes.charAt(i),count[i],

            new String(new char[count[i]]).replace('\0', '*')));

Answer 1

Your algorithm isn't bad, you could maybe speed things up a bit by using the ASCII values of the characters to directly lookup into the array and save the indexOf step.

ie c-'A' is a number from 0 to 26 for upper case characters.

At the moment your program will fall over if it hits any characters not in the mapping array.

To solve that you could consider using a Map from char to Integer to store the counts, adding characters to the key of the map with a count value of 1 as you find them, incrementing the count by one if it is already present.

Answer 2

import java.util.ArrayList;
import java.util.Collections;
import java.util.List;

public class MyPrg {
    public static void main(String[] args) {
        String name = "afacfa";

        System.out.println("Input Text = " + name);
        System.out.println("Result = " + arrangeCharactersInIncreasingOrderOfFrequency(name));

    }

    public static String arrangeCharactersInIncreasingOrderOfFrequency(String inputString) {
        String result = "";
        List<String> updatedStringList = new ArrayList<String>();

        for (int i = 0; i < inputString.length(); i++) {
            updatedStringList.add(inputString.substring(i, i + 1));
        }
        Collections.sort(updatedStringList);

        for (String abc : updatedStringList) {
            result = result.concat(abc);
        }
        return result;
    }
}

Answer 3

If you are familiar with hash maps, this code will easily do what you want. The hashmap value is the frequency counter and it prints the output by finding the max value inside the hashmap.

import java.util.Arrays;
import java.util.HashMap;

public class FrequencyPrint {
    public static void main(String[] args) {
        String s = "ccrrcdcffcghijk";
        HashMap<Character, Integer> hashMap = new HashMap<Character, Integer>();
        for (int i = 0; i < s.length(); i++) {
            if (hashMap.containsKey(s.charAt(i))) {
                int value = hashMap.get(s.charAt(i));
                hashMap.put(s.charAt(i), ++value);
            } else {
                hashMap.put(s.charAt(i), 1);
            }
        }

        Character keys[] = Arrays.copyOf(hashMap.keySet().toArray(), hashMap
                .keySet().toArray().length, Character[].class);
        Integer values[] = Arrays.copyOf(hashMap.values().toArray(), hashMap
                .values().toArray().length, Integer[].class);


        for (int i = 0; i < keys.length; i++) {
            int x = FrequencyPrint.findmax(values);
            for (int j = 0; j < values[x]; j++) {
                System.out.print(keys[x]);
            }
            values[x] = 0;
        }

    }

    public static int findmax(Integer values[]) {
        int max = 0;
        for (int i = 0; i < values.length; i++) {
            if (values[i] > values[max]) {
                max = i;
            }
        }
        return max;
    }
}