在Java使用HashMap将二进制数转换为八进制数

请记住，当我们将八进制转换为十进制时，我们一次获取了3个二进制数字。将使用类似的方法，在这里，对于每3位数字，我们有一个对应的数字，如在八进制系统中，我们有从0到’R-1’的数字，其中R代表数字系统的基值。顾名思义，在八进制系统中，R等于8。因此，该数字如下：0、1、2、3、4、5、6、7。

现在，借助HashMaps转换一件东西的过程很明显，一种方式或其他二进制等价物和八进制等价物将在我们的HashMap中充当键值对。

Octal	Binary
0	000
1	001
2	010
3	011
4	100
5	101
6	110
7	111

算法：

1. 将二进制数转换为十进制数。
2. 使用HashMap，我们将每个位及其十进制值进行映射。
3. 找到该位时，它将与十进制数字对应
4. 打印并显示该数字的八进制等效项。

插图：

Input  1 : 1011
Output 1 :   13

Input  2 : 1000
Output 2 :   10

执行：

例子

Java

// Java Progeram to Convert Binary Number to Octal Number
  
// Importing all utility classes
import java.util.*;
  
// Main class
pubic class GFG {
  
    // Mehod 1
    // Helper method
    public static void octal(String s)
    {
  
        // Here binary number is representedd by string 's'
        // over which standard length() method is computed
        // to get the length of binary number
  
        // Appending 2 zeros if bnary numbers leaves
        // remainder as 1 after dividing with 3
        if (s.length() % 3 == 1) {
  
            // Append  two zeros to it
            s = "00" + s;
        }
  
        // If binary string number length after equals 2
        else if (s.length() % 3 == 2) {
  
            // Concatinate string by adding 1 zero to it
            s = "0" + s;
        }
  
        // Creating an object of HashMap
        // Declaring object of String and Integer types
        HashMap hm = new HashMap<>();
  
        // Adding elements to the object created above
        // using the put() method
  
        // Adding elements(key-value) pairs to given object
        // 000 in binary system  ->  0 in octal system
        // 001 in binary system  ->  1 in octal system
  
        // Similarly adding for 0 to N-1 (N=8 for octal)
        hm.put("000", 0);
        hm.put("001", 1);
        hm.put("010", 2);
        hm.put("011", 3);
        hm.put("100", 4);
        hm.put("101", 5);
        hm.put("110", 6);
        hm.put("111", 7);
  
        // Creating and declaring a string array
        String[] part = new String[3];
        int t = 0;
  
        // Iterating over the binary number digits
        for (int i = 0; i < s.length(); i = i + 3) {
  
            // Checking for substring in an binary number
            // gigit array
            String bypart = s.substring(i, i + 3);
            part[t] = bypart;
  
            // If found
            if (hm.containsKey(part[t])) {
  
                // Getting the part to be matched for
                // its corresponding octal numbers
                System.out.print(hm.get(part[t]));
            }
  
            // Incrementing the counter
            t++;
        }
    }
  
    // Method 2
    // Main driver method
    public static void main(String[] args)
    {
  
        // Display message 
        System.out.print("Enter the binary number to be converted : ");
        // Binary number to be converted
        // Custom entry  
        String s = 011;
          
        // Calling the method1 octal() over the
        // above input entered number
        octal(s);
        
        // Display message 
        System.out.print("Octal equivalent : ");
    }
}

输出：

Enter the binary number to be converted : 011
Octal equivalent : 3