如何在java中捕获ascii中缺少的前导零

我创建了一个程序，通过

扫描仪

输入接收字符串，并将其转换为二进制ASCII数字，还打印出每个字母的频率。 这是我的程序

import java.util.*;

public class ASCII
{
    public static void main (String [] args)
    {
        Scanner sc = new Scanner (System.in);
        System.out.println("Enter your sentence: " );
        String myString = sc.nextLine();
        int length = myString.length();
        int ascii;
        char character;
        int[] myarray = new int[256];
        for(int i =0; i < length; i++)
        {
            character = myString.charAt(i);
            ascii = (int)character; // casting the character into an int
            System.out.print(Integer.toBinaryString(ascii) + " ");
            myarray[ascii]++; // counting the frequency of each letter in the string
        }

        System.out.println();

        for ( int k = 0; k < myarray.length; k++ )
        {
            if ( myarray[k] > 0 ) // if the frequency is greater than 0 do this print line, without this programm would print all 255 ascii characters.
            {
                System.out.println("'"+(char)k + "'" + " appeared " +    myarray[k]      + " times." );
            }
        }
    }
}

但我得到了：

"1101000 1100101 1101100 1101100 1101111 100000 1110100 1101000 1100101 1110010 1100101"

正如您所看到的，“100000”只有6位长，我的一些ascii值由于缺少前导零的问题而错误

我不知道我将如何着手解决这个问题，这就是为什么我在这里。如果有人能告诉我如何解决这个问题，那就太好了。

publicstaticvoidmain（String[]args）
public static void main (String [] args)
{
    String myString = "hello there";
    int length = myString.length();
    int ascii;
    char character;
    int[] myarray = new int[256];
    for(int i =0; i < length; i++)
    {
        character = myString.charAt(i);
        ascii = character; // casting the character into an int
        // not optimal, but works
        System.out.println(String.format("%7s", Integer.toBinaryString(ascii)).replace(' ', '0'));

        myarray[ascii]++; // counting the frequency of each letter in the string

    }
    System.out.println();

    for ( int k = 0; k < myarray.length; k++ )
    {
        if ( myarray[k] > 0 ) // if the frequency is greater than 0 do this print line, without this programm would print all 255 ascii characters.
        {
            System.out.println("'"+(char)k + "'" + " appeared " +    myarray[k]      + " times." );
        }
    }
}

{
String myString=“你好”；
int length=myString.length（）；
int-ascii；
字符；
int[]myarray=新int[256]；
for（int i=0；i0）//如果频率大于0，则执行此打印行，如果没有此程序，则将打印所有255个ascii字符。
{
System.out.println（“'”+（char）k+“'+”出现“+myarray[k]+”次”）；
}
}
}

您应该停止将字符串转换为整数并直接比较字符串，或者将k转换为字符串，附加所需数量的前导零，然后将其转换为字符。

使用时发生的情况：System.out.print（Integer.toString（ascii，2）+“”）；是的，有128个ASCII代码点，编号为0-127，编码为值0-127。但是，基于Java使用ASCII的假设，您的程序中有一些坚持。Java、JavaScript、.NET、XML、HTML……使用

charAt（i）

返回一个UTF-16编码单元，其中一个或两个编码一个Unicode码点，例如a或欧元，或者使用这种方法，它实际上每个字母打印8位，而不是7位。您能修复它，使其打印7位吗？

public static void main (String [] args)
{
    String myString = "hello there";
    int length = myString.length();
    int ascii;
    char character;
    int[] myarray = new int[256];
    for(int i =0; i < length; i++)
    {
        character = myString.charAt(i);
        ascii = character; // casting the character into an int
        // not optimal, but works
        System.out.println(String.format("%7s", Integer.toBinaryString(ascii)).replace(' ', '0'));

        myarray[ascii]++; // counting the frequency of each letter in the string

    }
    System.out.println();

    for ( int k = 0; k < myarray.length; k++ )
    {
        if ( myarray[k] > 0 ) // if the frequency is greater than 0 do this print line, without this programm would print all 255 ascii characters.
        {
            System.out.println("'"+(char)k + "'" + " appeared " +    myarray[k]      + " times." );
        }
    }
}