在哈希图中存储字符和二进制数

Question

我正在尝试存储字母到二进制数的映射。这是我的映射

("h",001)
("i", 010)
("k",011)
("l",100)
("r", 101)
("s",110)
("t",111)

为此，我创建了一个哈希映射并存储了键值对。我现在想为给定的句子显示相应的二进制值。这是我的代码。

package crups;

import java.util.*; 


public class onetimepad {

public static void main(String args[])
{
    HashMap <String , Integer>hm  = new HashMap <String , Integer> (); 
    hm.put("e", 000);
    hm.put("h",001);
    hm.put("i", 010);
    hm.put("k",011);
    hm.put("l",100);
    hm.put("r", 101);
    hm.put("s",110);
    hm.put("t",111);



    String[] key = { "t" ,"r" , "s" , "r","t","l","e", "r","s","e"};
    //key = t r s r t l e r s e
    String[] input = {"h","e","i" ,"l","h","i","t","l","e","r"};
    int[] cipher = new int[10]; 
    System.out.println("Binary form of text is ....");

    for( String s : input )
    {
        System.out.print(hm.get(s)+" ");
    }

}   

}

但是，当我运行代码时，字母“i”的映射显示错误：8：而不是010。 有人可以告诉我为什么会这样吗？另外，我如何在我的数字前面显示零，因为这些是二进制数。 谢谢。

输出：

Binary form of text is ....
1 0 8 100 1 8 111 100 0 101 

Answer 1

感谢您的投入。 我已经完成了我的问题的一部分，这需要我加密我的输入文本：“Heilhitler”并以二进制格式显示它。这是我根据您的建议构建的代码。

public static void main(String args[])
    {
        HashMap <String , Integer>hm  = new HashMap <String , Integer> (); 
        hm.put("e", 0);
        hm.put("h",1);
        hm.put("i", 2);
        hm.put("k",3);
        hm.put("l",4);
        hm.put("r",5);
        hm.put("s",6);
        hm.put("t",7);
String[] key = { "t" ,"r" , "s" , "r","t","l","e", "r","s","e"};
    //key = t r s r t l e r s e
    String[] input = {"h","e","i" ,"l","h","i","t","l","e","r"};

    int[] inter1 = new int[10];     
    System.out.println("Binary form of text is ....");
    int i = 0 ; 
    for( String s : input )
    { 

        String binarystr = Integer.toBinaryString(hm.get(s)) ; 
        System.out.print( binarystr+" ");
        inter1[i]=Integer.parseInt(binarystr) ; 
        i++ ; 
    }

    int[] inter2 = new int[10]; 
    int m= 0 ; 
    for( String s : key )
    { 
        String binarystr = Integer.toBinaryString(hm.get(s)) ; 
        System.out.print( binarystr+" ");
        inter2[m]=Integer.parseInt(binarystr) ; 

        m++ ; 
    }


    int[] cipher = new int[10];
    for(int j = 0 ; j < 10 ; j++)
    {
        cipher[j] = inter1[j] ^ inter2 [j];   //performing xor between input and key 
    }
    System.out.println("Cipher is .....");
    for( int j= 0 ; j < 10;  j++ )
    {
        System.out.print(" " + cipher[j]);
    }

   //-------------------Decryption //----------------------------
    //Performing XOR between the cipher and key again 
    int[] decry = new int[10] ; 

    for(int j = 0 ; j < 10 ; j ++ )
    {
        decry[j] = cipher[j] ^ inter2[j]; 
    }
    System.out.println(" ");
    System.out.println("Decrypted result in Binary format");
    for( int j= 0 ; j < 10;  j++ )
    {
        System.out.print(" " + decry[j]);
    }

    }
  }

输出：

Binary form of text is ....
1 0 10 100 1 10 111 100 0 101  
Cipher is .....
110 101 100 1 110 110 111 1 110 101 
Decrypted result in Binary format
1 0 10 100 1 10 111 100 0 101

Answer 2

你不能用前导零来存储它们。前导零到整数表示它是八进制数。

由于下一步是异或，我推荐这种方法。

1. 您可以使用简单的以 10 为基数的数字来存储这些整数。我们将在需要时将它们转换为二进制文件。 （您也可以将它们简单地存储为带有前导 0b 的二进制文件。See this answer for more details.
2. 使用Integer.toString(hm.get(s), 2); 显示二进制数。原始数字仍然是整数，因此您可以将其用于异或运算。

3. 为了显示带有前导零的二进制文件，我使用了一些这样的字符串方法：

   String temp = "000", binary;
   for( String s : input ) {
       binary = Integer.toString(hm.get(s), 2);
       System.out.print(temp.substring(0, 3-binary.length()) + binary +" ");
   }
   

这是最终代码的样子：

输出

Binary form of text is ....
001 000 010 100 001 010 111 100 000 101 

代码

import java.util.*;

public class onetimepad {
    public static void main(String args[]) {
        HashMap <String , Integer>hm  = new HashMap <String , Integer> (); 
        hm.put("e", 0); //or use hm.put("e", 0b000);
        hm.put("h", 1); //or use hm.put("e", 0b001);
        hm.put("i", 2);
        hm.put("k", 3);
        hm.put("l", 4);
        hm.put("r", 5);
        hm.put("s", 6);
        hm.put("t", 7);

        String[] key = { "t" ,"r" , "s" , "r","t","l","e", "r","s","e"};
        //key = t r s r t l e r s e
        String[] input = {"h","e","i" ,"l","h","i","t","l","e","r"};
        int[] cipher = new int[10]; 
        System.out.println("Binary form of text is ....");

        String temp = "000", binary;
        for( String s : input ) {
            binary = Integer.toString(hm.get(s), 2);
            System.out.print(temp.substring(0, 3-binary.length()) + binary +" ");
        }
    }   
}

Answer 3

首先，您的Map 声明和初始化有点偏离。要使用二进制常量，请在其前面加上 0b - 并请编程到 Map 接口（不是 HashMap 实现）。而且，从 Java 7 开始，您可以使用菱形运算符 <> 来缩短时间。

Map<String, Integer> hm = new HashMap<>();
hm.put("e", 0b000);
hm.put("h", 0b001);
hm.put("i", 0b010);
hm.put("k", 0b011);
hm.put("l", 0b100);
hm.put("r", 0b101);
hm.put("s", 0b110);
hm.put("t", 0b111);

然后，对于打印，您有 Integer(s) 但您想要它们的二进制表示。所以你可以做类似的事情，

for (String s : input) {
    System.out.print(Integer.toBinaryString(hm.get(s)) + " ");
}

我跑去拿的（我相信你已经预料到了）

Binary form of text is ....
1 0 10 100 1 10 111 100 0 101 

如果您真的想要前导零（三位二进制格式），您可以这样做，

for (String s : input) {
    StringBuilder sb = new StringBuilder(Integer.toBinaryString(hm.get(s)));
    while (sb.length() < 3) {
        sb.insert(0, '0');
    }
    System.out.print(sb.append(" "));
}

哪些输出

Binary form of text is ....
001 000 010 100 001 010 111 100 000 101