我的代码中查找字符频率有什么问题？

Question

我正在尝试输出字符串中的所有字母及其频率，但只有a 出现。我认为我的逻辑是正确的，但看起来我遗漏了一些东西。

我做错了什么，我该如何解决？

这是我的代码：

public static void solution(String s) {
    char[] c = s.toCharArray();

    int j = 0, i = 0, counter = 0;

    for(i = 1; i < c.length; i++) {
        if(c[i] != c[j]) {
            i++;
        } else {
            counter++;
        }
    }
    System.out.println("The letter " + c[j] + " appears " + counter + " times");
}

public static void main(String args[]) {
    String s = "abaababcdelkm";
    solution(s);
}

输出：

The letter a appears 1 times

Answer 1

每次通过循环时，都会将 i 递增两次。因此，您只是在测试其他所有角色。在循环内部使用 continue 而不是 i++。

public static void solution(String s) {
    char[] c = s.toCharArray();

    int j = 0, i = 0, counter = 0;

    for(i = 1; i < c.length; i++) {
        if(c[i] != c[j]) {
            continue;
        } else {
            counter++;
        }
    }
    System.out.println("The letter " + c[j] + " appears " + counter + " times");
}

请注意，此代码将告诉您字符串中的第一个字符出现在字符串的其余部分中的次数。也许这就是您想要的，但您的问题并不清楚。

Answer 2

您的代码中没有任何预防措施可以使代码在功能步骤中不计算相同的字符。在这里，我刚刚修改了您的代码以使其正常工作。但是您可以将其与我提供的其他版本进行比较，以防止重复计算。

public class Main {

public static void solution(String s) {
    char[] c = s.toCharArray();
            int j = 0, i = 0, counter = 0;
    for (i = 0; i < c.length; i++) {
        for (j = i; j < c.length; j++) {

            if (c[i] == c[j]) {
                counter++;

            }
        }
        System.out.println("The letter " + c[i] + " appears " + counter + " times");
        counter = 0;
    }
}

public static void main(String args[]) {
    String s = "abaababcdelkm";
    solution(s);
}
}

输出：

The letter a appears 4 times
The letter b appears 3 times
The letter a appears 3 times
The letter a appears 2 times
The letter b appears 2 times
The letter a appears 1 times
The letter b appears 1 times
The letter c appears 1 times
The letter d appears 1 times
The letter e appears 1 times
The letter l appears 1 times
The letter k appears 1 times
The letter m appears 1 times

Answer 3

我添加了 Arraylist 来跟踪已经计数的字符，每个字符只会打印一次

import java.util.ArrayList;

public class Main {

    public static void solution(String s) {
        char[] c = s.toCharArray();
        ArrayList<Character> arrayList;
        arrayList = new ArrayList<Character>();

        int j = 0, i = 0, counter = 0;

        for (i = 0; i < c.length; i++) {
            if (!arrayList.contains(c[i])) {
                arrayList.add(c[i]);
                for (j = i; j < c.length; j++) {

                    if (c[i] == c[j]) {
                        counter++;
                    }
                }
                System.out.println("The letter " + c[i] + " appears " + counter + " times");
                counter = 0;
            } else {
                continue;
            }

        }

    }

    public static void main(String args[]) {
        String s = "abaababcdelkm";
        solution(s);
    }
}

输出：

The letter a appears 4 times
The letter b appears 3 times
The letter c appears 1 times
The letter d appears 1 times
The letter e appears 1 times
The letter l appears 1 times
The letter k appears 1 times
The letter m appears 1 times

Answer 4

这是使用地图的一种有效方式

import java.util.Map;
import java.util.Scanner;
import java.util.TreeMap;


public class Main {

    public static void main(String[] args) {
        String s = "abaababcdelkm";
        Map<Character,Integer> frequency = new TreeMap<Character,Integer>();


        for (int i = 0 ; i< s.length() ; i++) {
            char ch = s.charAt(i);

            if (frequency.containsKey(ch)) {

                frequency.replace(ch, frequency.get(ch) + 1);

            }
            else {
                frequency.put(ch, 1);

            }

        }

        for (Map.Entry<Character,Integer> entry : frequency.entrySet()) {
            System.out.println("The letter " + entry.getKey() + " appears " + entry.getValue() + " times");


        }
    }

}