Question

我试图让我的程序在java中计算每个单词中的字母数。现在，我有它计算单词，而不是字母。任何帮助它来做信都会很棒！

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

public class LetterTypeCount 
{
    public static void main(String[] args) {
        // create HashMap to store String keys and Integer values
        Map<String, Integer> myMap = new HashMap<>();

        createMap(myMap); // create map based on user input
        displayMap(myMap); // display map content
    }

    // create map from user input
    private static void createMap(Map<String, Integer> map)
    {
        Scanner scanner = new Scanner(System.in); // create scanner
        System.out.println("Enter a string:"); // prompt for user input
        String input = scanner.nextLine();

        // tokenize the input
        String[] tokens = input.split(" ");

        // processing input text
        for (String token : tokens)
        {
            String letter = token.toLowerCase(); // get lowercase letter

            // if the map contains the letter
            if (map.containsKey(letter)) // is letter in map
            {
                int count = map.get(letter); // get current count
                map.put(letter, count + 1); // increment count
            }
            else
                map.put(letter, 1); // add new letter with a count of 1 to map
        }
    }

    // display map content
    private static void displayMap(Map<String, Integer> map)
    {
        Set<String> keys = map.keySet(); // get keys

        // sort keys
        TreeSet<String> sortedKeys = new TreeSet<>(keys);

        System.out.printf("%nMap contains:%nKey\t\tValue%n");

        // generate output for each key in map
        for (String key : sortedKeys)
            System.out.printf("%-10s%10s%n", key, map.get(key));
        System.out.printf("%nsize: %d%nisEmpty: %b%n",
                map.size(), map.isEmpty());
    }
} // end class LetterTypeCount

我想我需要在某处使用String charAt方法

Answer 1

可能不是很优雅但是：

public class LetterTypeCount 
{
    public static void main(String[] args) {
        // create HashMap to store String keys and Integer values
        Map<String, Integer> myMap = new HashMap<>();

        createMap(myMap); // create map based on user input
        displayMap(myMap); // display map content
    }

    // create map from user input
    private static void createMap(Map<String, Integer> map)
    {
        Scanner scanner = new Scanner(System.in); // create scanner
        System.out.println("Enter a string:"); // prompt for user input
        String input = scanner.nextLine();

        // split to words
        String[] words = input.split(" ");
        for (String word : words) 
        {
            word = word.toLowerCase(); // get lowercase word
            for(int i=0; i<word.length(); i++) 
            {
                char c = word.charAt(i); //get char at position i
                if (map.containsKey(c + "")) // is letter in map
                {
                    int count = map.get(c + ""); // get current count
                    map.put(c + "", count + 1); // increment count
                }
                else
                    map.put(c + "", 1); // add new letter with a count of 1 to map
            }
            // if the map contains the letter
        }
    }

    // display map content
    private static void displayMap(Map<String, Integer> map)
    {
        Set<String> keys = map.keySet(); // get keys

        // sort keys
        TreeSet<String> sortedKeys = new TreeSet<>(keys);

        System.out.printf("%nMap contains:%nKey\t\tValue%n");

        // generate output for each key in map
        for (String key : sortedKeys)
            System.out.printf("%-10s%10s%n", key, map.get(key));
        System.out.printf("%nsize: %d%nisEmpty: %b%n",
                map.size(), map.isEmpty());
    }
} // end class LetterTypeCount

Answer 2

假设你只计算拉丁字符：

public int[] letters = new int['z' - 'a' + 1];

// count chars:
for (String token : tokens) 
    for (char c : token.toLowerCase().toCharArray()) 
        if (c >= 'a' && c <= 'z')
            letters[c - 'a']++;

// then print out counts:
for (int i = 0; i < 'z' - 'a' + 1; i++) 
    if (letters[i] > 0)
        System.out.format("%s occurs %s time(s)\n", (char)('a' + i), letters[i]);

这看起来有点“低级”，但这是有效的Java代码（刚刚测试过！），它使用char和int s在算术运算上的可互换性。

Answer 3

您可以创建一个for循环，并为每个字母添加+1给数组或包含字母出现的HashMap。

for(int i=0; i<= input.length(); i++){
mymap.put(input.charAt(i), myMap.get(input.charAt(i)+1));
}

但是你应该定义你的变量myMap全局而不是你的main方法。

Answer 4

您可以尝试Java 8 Stream API
除了Java流之外，您还应该了解Java Lambda Expressions

String input = "ANY String";
    Map<String, Long> map = Arrays.stream(input.split("")) // Stream String
            .map(String::toLowerCase) // All letters to lower case
            .filter(letter -> !letter.equals(" ")) // Remove spaces
            .collect(Collectors.groupingBy(Function.identity(), Collectors.counting()));

Answer 5

你可以像这样简单地使用str.length()：

System.out.println(key.length());

计算字母出现的字母

5 个答案: