Java中严格使用乘法和除法将二进制数转换为十进制数

Question

我正试图让我的二进制到十进制方法正确计算。我必须使用乘法和/或除法在"bToD"方法。我不知道如何让它返回正确的答案。它应该只返回"5"，但是它返回了"45"。我需要修复这个方法，以便继续剩余的十六进制到二进制和八进制到二进制方法。

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

这是一个解决方案，你的解决方案相当复杂。

//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);
}