严格使用乘法和除法将 Java 中的二进制数转换为十进制数

Question

我正在尝试让我的二进制到十进制方法正确计算。我必须在“bToD”方法中使用 multiply and/or divide 方法。我不知道如何得到它 return 正确的答案。它应该只是 returning "5"，但它是 returning 45。我需要修复此方法以继续剩余的十六进制到二进制和八进制到二进制的方法。

package numType;

import java.util.Scanner;


public class  numType{

    public static String multiply2(String n) { //return 2*n
        String r = "";
        int c = 0;
        for (int i = n.length()-1; i >= 0; i--) {
            int p = (n.charAt(i)-'0')*2+c;
            c = p/10;
            r = p%10+r;
        }
        if (c > 0)
            r = c+r;
        return r;
        }
    
    public static String divide2(String n) { //return n/2
        String r = "";
        int b = 0;
        int i = 0;
        if (n.charAt(0) < '2') {
            b = 1;
            i = 1;
        }
        for (; i < n.length(); i++) {
            int p = (n.charAt(i)-'0')+b*10;
            b = p%2;
            r += p/2;
        }
        if (r.length() == 0)
            r = "0";
        return r;
    }
    
    //convert binary string b to an equivalent decimal string.
    public static String bToD(String b)
    {
        String s = "";
        int n = 0;
        if (b.charAt(b.length()-1) == '1')
            n = 1;
        else
            n = 0;
        int pow = 1;                            //INITIALIZE POWER # FROM THE SECOND TO LAST 2^1
        for (int i=b.length()-2; i>=0; i--)    // LIST #'S FROM MOST RIGHT-HAND SIDE
        {
            char ch = b.charAt(i);
            String temp = "" + ch;
            for (int j=1; j<=pow; j++) 
                temp = multiply2(temp);
            //System.out.println(temp);
            int n1 = 0;
            for (int k=0; k<temp.length(); k++)
            {
                n1 = n1*10 + (int) (temp.charAt(k)-'0');
            }
            n = n + n1;
            s = temp;
            pow++;
        }
        s = s + n;
        return s;
    }
    
    //convert decimal string d to an equivalent binary string.
    public static String dToB(String d)
    {
        String s = "";
        while (!d.equals("0"))
        {
            String d1 = divide2(d);
            d1 = multiply2(d1);
            if (d1.equals(d))
                s = "0" + s;
            else
                s = "1" + s;
            d = divide2(d);
        }
        return s;
    }
    
    //convert binary string b to an equivalent octal string.
    public static String bToO(String b) 
    {
        String s = "";
        int groups = b.length()/3;
        int index = 0;
        //System.out.println(index);                       //bToD(b)
        while (groups != index)
        {
            for (int i = b.length()-3; i >= 0; i--) 
            {
                for (int j=1; j<=i; j++)
                {
                    String temp = b.substring(b.length()-3,b.length()); //last 3 digits in binary
                    String sSub = b.substring(0,b.length()-3);          //first digits in binary
                    s = bToD(temp);
                    

                    
                }
            }
            index++;
        }
        return s;
        
    }
    
    //convert octal string o to an equivalent binary string.
    public static String oToB(String o) 
    {
        String s ="";
        int digits = o.length();
        int index = 0;
        while (digits != index)
        {
            for (int i=o.length()-1; i>=0; i--)
            {
                char ch = o.charAt(i);
                //System.out.println(digits);
                
                switch (ch)
                {
                    case '7':
                        s = s + "111";
                        index++;
                        break;
                    case '6':
                        s = s + "110";
                        index++;
                        break;
                    case '5':
                        s = s + "101";
                        index++;
                        break;
                    case '4':
                        s = s + "100";
                        index++;
                        break; 
                    case '3':
                        s = s + "011";
                        index++;
                        break;   
                    case '2':
                        s = s + "010";
                        index++;
                        break; 
                    case '1':
                        s = s + "001";
                        index++;
                        break;   
                    case '0':
                        s = s + "000";
                        index++;
                        break;    
                }
            }
        }
        return s;
    }
    
    //convert binary string b to an equivalent hexadecimal string.
    public static String bToH(String b) 
    {
        String s ="";
        return s;
    }
    
    //convert hexadecimal string h to an equivalent binary string.
    public static String hToB(String h) 
    {
        String s ="";
        return s;
    }
    
    public static void main(String[] args) {
        // TODO code application logic here
        String b,d,o,h;
        
        b = "101";
        System.out.println("Binary to Decimal:");
        System.out.println(b + " => " + bToD(b));
        
        System.out.println();
        
        System.out.println("Decimal to Binary:");
        d = "45";
        System.out.println(d + " => " + dToB(d));
        
        System.out.println();
        
        System.out.println("Binary to Octal:");
        b = "100101101";
        System.out.println(b + " => " + bToO(b));
        
        System.out.println();
        
        System.out.println("Octal to Binary:");
        o = "";
        System.out.println(o + " => " + oToB(o));
        
        System.out.println();
        
        System.out.println("Binary to Hexadecimal:");
        b = "";
        System.out.println(b + " => " + bToH(b));
        
        System.out.println();
        
        System.out.println("Hexadecimal to Binary:");
        h = "";
        System.out.println(h + " => " + hToB(h));
        
        }
    }

Answer 1

这里有一个解决方案，你的解决方案很复杂。

    //convert binary string b to an equivalent decimal string.
    public static String bToD(String b)
    {
        int ans = 0, pow = 0;
        
        //For every digit in binary
        for(int i=b.length()-1; i>=0; i--){
            // Get string of current char
            String cur = Character.toString(b.charAt(i));
            
            for (int j=0; j<pow; j++) 
                cur = multiply2(cur);
            
            ans += Integer.parseInt(cur);
            pow++;
        }
        
        return Integer.toString(ans);
    }

严格使用乘法和除法将 Java 中的二进制数转换为十进制数

Converting Binary to Decimal Numbers in Java strictly using multiplication and division

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