Question

我需要多行文本并计算每行中每个字母的出现次数。最后一行必须在一段时间内结束。我要求在arrayName[0] = number of a's，arrayName[1] = number of b's等处创建一个长度为26的int数组，并且需要忽略该字母大小写。我无法检查每个字母的出现次数，并使用正确的出现次数定义索引变量。到目前为止我的代码：

import java.util.Scanner;

public class LetterCounter 
{
public static void main(String[] args)
{
    Scanner input = new Scanner(System.in);

    String[] textBlock = new String[10];

    System.out.println("Enter text.");
    int index;
    //Allows me to type in something for all "lines" of text until I end one with a period.
    for(index = 0; index < textBlock.length; index++)
    {
        textBlock[index] = input.nextLine();

        if(textBlock[index].contains("."))
        {
            System.out.println("Period found");
            break;
        }
    }
    //Outputs what the user printed except for the lines not typed in (they return null).
    System.out.println("Text input by user:");
    for(index = 0; index < textBlock.length; index++)
    {
        if(textBlock[index] != null)
        {
            System.out.println(textBlock[index]);
        }
    }
    //int array
    int[] occurrences = new int[26];

    //used to turn letter into number
    char letter = 'a' - 49;
    char letter2 = 'b' - 49;
    char letter3 = 'c' - 49;

    //checks that numbers are correct (return "0" for a, "1" for b, and "2" for c)
    System.out.println("Number value of a: " + letter);
    System.out.println("Number value of b: " + letter2);
    System.out.println("Number value of c: " + letter3);

}// End of main
}//End of program

Answer 1

计算每一行的字符出现次数。然后打印它们。试试这个...

//final int SIZE = 1000;
//String[] textBlock = new String[SIZE];

Scanner in = new Scanner(System.in);
String line;

//frequency array for storing which Character is occurs how many times
int[] frequencyArray = new int[26];
Arrays.fill(frequencyArray, 0);

System.out.println("Enter text : ");

// take input un-till find the (.) in a line
// and also count the frequency of Character of current line
while (true) {
    line = in.nextLine();
    for (int i = 0; i < line.length(); i++) {
        char ch = Character.toLowerCase(line.charAt(i));
        if (Character.isLetter(ch)) {
            frequencyArray[ch - 'a']++;
        }
    }
    if (line.contains(".")) {
        break;
    }
}

for (int i = 0; i < 26; i++) {
    System.out.println("Total Number of " + (char)(i + 'a') + " : " + frequencyArray[i]);
}

Answer 2

在循环外声明数组，并在字符匹配时增加该数组的计数

在循环中声明这个

int intialCount = occurrences(charAt(textBlock[index])-49);
occurrences(charAt(textBlock[index])-49) = intialCount++;

你会发现＆＃39; a＆＃39; in occurences [0]将返回一个计数。

Answer 3

如果您不打印调试消息，则只能使用1行进行操作：

System.out.println("Enter lines of text (period to end):");
new Scanner(System.in).useDelimiter("\\.").next().chars()
  .filter(i -> i >= 'a' && i <= 'c').boxed()
  .collect(Collectors.groupingBy(i -> i, Collectors.counting())
  .forEach((c, n) -> System.out.format("Number value of %s: %d\n", (char)c, n));

java - 将字符串中的字母出现计数为26的int数组

3 个答案: