计算字符串的字母(大写和小写)

时间:2012-07-18 14:35:35

标签: java file-io for-loop

我这里有一个程序,它输入一个段落并将其写入文件。之后,它应计算每个字母的出现次数(区分大小写)。但是,它不计算字母出现次数。我想我把 for循环放在了错误的地方。

import java.io.*;
import java.util.*;
public class Exercise1 {

    public static int countLetters (String line, char alphabet) {
        int count = 0;
        for (int i = 0; i <= line.length()-1; i++) {
            if (line.charAt(i) == alphabet)
                count++;
        }
        return count;
    }

    public static void main(String[] args) throws IOException {
        BufferedReader buffer = new BufferedReader (new InputStreamReader(System.in));
        PrintWriter outputStream = null;
        Scanner input = new Scanner (System.in);
        int total;

        try {
            outputStream = new PrintWriter (new FileOutputStream ("par.txt"));
            System.out.println("How many lines are there in the paragraph you'll enter?");
            int lines = input.nextInt();
            System.out.println("Enter the paragraph: ");
            String paragraph = buffer.readLine();
            outputStream.println(paragraph);
            int j;
            for (j = 1; j<lines; j++) {
                paragraph = buffer.readLine();
                outputStream.println(paragraph);    
            }
            outputStream.close();
            System.out.println("The paragraph is written to par.txt");

            for (int k=1; k<lines; k++) {
                paragraph = buffer.readLine();
                total = countLetters (paragraph, 'A');
                if (total != 0)
                    System.out.println("A: "+total);
                            //I'll do bruteforce here up to lowercase z

            }
        }

        catch(FileNotFoundException e) {
            System.out.println("Error opening the file par.txt");
        }

    }

}

请帮我修复代码。我是编程新手,我需要帮助。非常感谢你!

4 个答案:

答案 0 :(得分:6)

首先,你的初始阅读用户输入有点浪费,因为你读了一次然后进入for循环的其余部分 - 这不是问题,只是一个更好的代码。

// your code
String paragraph = buffer.readLine();
outputStream.println(paragraph);
int j;
for (j = 1; j<lines; j++) {
     paragraph = buffer.readLine();
     outputStream.println(paragraph);    
 }

你可以把它们放在循环中:

// better code
String paragraph;
int j;
for (j = 0; j<lines; j++) {
     paragraph = buffer.readLine();
     outputStream.println(paragraph);    
}

然后你的第一个问题来自你读行的方式:

// your code - not working
outputStream.close();
for (int k=1; k<lines; k++) {
      paragraph = buffer.readLine();
      total = countLetters (paragraph, 'A');

考虑上面发生的事情:

  • 输入已经 DONE ,输出已经写入并且流已关闭 - 到此为止一切都很好
  • 然后,当您尝试计算字符数时,您会执行:paragraph = buffer.readLine(); - 此代码的作用是什么?它等待另一个用户输入(而不是读取已插入的内容)

要解决上述问题:您需要阅读已经已写入的内容 - 而不是要求其他输入。然后,您可以将它们放入列表并编写for循环,而不是逐个强制逐个强制每个字符。

现在,您想要读取您已经创建的现有文件(即读取用户输入的内容):

BufferedReader fileReader = new BufferedReader(new FileReader(new File("par.txt")));

String allCharacters = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ";
String aLineInFile;

// Read the file that was written earlier (whose content comes from user input)
// This while loop will go through line-by-line in the file
while((aLineInFile = fileReader.readLine()) != null)
{
      // For every line in the file, count number of occurrences of characters  
      // This loop goes through every character (a-z and A-Z)  
      for(int i = 0; i < allCharacters.length(); i++)
      {
            // For each single character, check the number of occurrences in the current line
            String charToLookAt = String.valueOf(allCharacters.charAt(i));
            int numOfCharOccurancesInLine = countLetters (aLineInFile, charToLookAt);

            System.out.println("For line: " + aLineInFile + ", Character: " + charToLookAt + " appears: " + numOfCharOccurancesInLine + " times " );
      }         
}

上面给出了每一行中每个字符的出现次数 - 现在你只需要对它们进行组织,以便跟踪整个文件的总数。

代码方面,可能有更好的方法来编写它以实现更清晰的实现,但上面的内容很容易理解(我只是很快就写了)。

答案 1 :(得分:0)

在一个循环中完成所有事情:

       for (j = 1; j<lines; j++) {
            paragraph = buffer.readLine();
            total = countLetters (paragraph, 'A');
            if (total != 0)
                System.out.println("A: "+total);
            outputStream.println(paragraph);    
        }

答案 2 :(得分:0)

您可以使用HashTable计算每个案例的致函:

        final Pattern patt = Pattern.compile("A-Za-z]");
        final HashMap<Character, Integer> tabChar = new HashMap<Character, Integer>(
            52);

        // replace : paragraph = buffer.readLine();
        // Unless you use it outside, you can declare it 'final'
        final char[] paragraph = "azera :;,\nApOUIQSaOOOF".toCharArray();


        for (final Character c : paragraph ) {
            if (Character.isLetter(c)) {
                Integer tot = tabChar.get(c);
                tabChar.put(c, (null == tot) ? 1 : ++tot);
            }
        }

输出:

{F=1, A=1, O=4, I=1, U=1, Q=1, S=1, e=1, a=3, r=1, p=1, z=1}

您可以使用final TreeSet<Character> ts = new TreeSet(tabChar.keySet());对字符进行排序,然后使用get(c);

tabChar对其进行排序

答案 3 :(得分:0)

之前的答案可以解决您的问题,但另一种避免暴力的方法可能是使用ASCII字符值循环。