将给定的BCD转换为等效的二进制格式

给定一个BCD(二进制编码的十进制) ，任务是将其转换为等效的二进制数。
例子：

Input: 1001000 
Output: 110000 
Explanation: 
Integer value of the given BCD is 48(0100 -> 4, 1000 -> 8). 
(48)₁₀ = (110000)₂
Input: 1001001 
Output: 110001 

方法：为了解决此问题，我们需要将给定的BCD编号拆分为长度为4的二进制块，并将它们一一转换为整数，以生成给定BCD的最终整数表示形式。生成后，将整数转换为其二进制形式。
下面是上述方法的实现。

C++

// C++ code to convert BCD to its
// equivalent Binary
 
#include 
using namespace std;
 
// Function to convert BCD to Decimal
string bcdToBinary(string s)
{
    int l = s.length();
    int num = 0;
    int mul = 1;
    int sum = 0;
     
     // If the length of given BCD is not
    // divisible by 4
    for (int i = l % 4 - 1; i >= 0; i--)
    {
        sum += (s[i] - '0') * mul;
        mul *= 2;
    }
    num = sum;
    sum = 0;
    mul = pow(2, 3);
    int ctr = 0;
    for (int i = l % 4; i < l; i++)
    {
        ctr++;
        sum += (s[i] - '0') * mul;
        mul /= 2;
        if (ctr == 4) {
            num = num * 10 + sum;
            sum = 0;
            mul = pow(2, 3);
            ctr = 0;
        }
    }
 
    // Convert decimal to binary
    string ans = "";
    while (num > 0) {
        ans += (char)(num % 2 + '0');
        num /= 2;
    }
    reverse(ans.begin(), ans.end());
    return ans;
}
 
// Driver Code
int main()
{
    string s = "1001000";
 
    // Function Call
    cout << bcdToBinary(s);
    return 0;
}

Java

// Java code to convert BCD to its
// equivalent Binary
import java.io.*;
import java.util.*;
 
class GFG{
     
// Function to convert BCD to Decimal
static String bcdToBinary(String s)
{
    int l = s.length();
    int num = 0;
    int mul = 1;
    int sum = 0;
     
    // If the length of given BCD is not
    // divisible by 4
    for(int i = l % 4 - 1; i >= 0; i--)
    {
        sum += (s.charAt(i) - '0') * mul;
        mul *= 2;
    }
 
    num = sum;
    sum = 0;
    mul = (int)Math.pow(2, 3);
    int ctr = 0;
     
    for(int i = l % 4; i < l; i++)
    {
        ctr++;
        sum += (s.charAt(i) - '0') * mul;
        mul /= 2;
         
        if (ctr == 4)
        {
            num = num * 10 + sum;
            sum = 0;
            mul = (int)Math.pow(2, 3);
            ctr = 0;
        }
    }
     
    // Convert decimal to binary
    String ans = "";
    while (num > 0)
    {
        ans += (char)(num % 2 + '0');
        num /= 2;
    }
 
    StringBuilder ans1 = new StringBuilder();
 
    // Append a string into StringBuilder input1
    ans1.append(ans);
 
    // Reverse StringBuilder input1
    ans = ans1.reverse().toString();
 
    return ans;
}
 
// Driver code
public static void main(String[] args)
{
    String s = "1001000";
     
    // Function call
    System.out.println(bcdToBinary(s));
}
}
 
// This code is contributed by coder001

Python3

# Python3 code to convert
# BCD to its equivalent Binary
 
# Function to convert
# BCD to Decimal
def bcdToBinary(s):
 
    l = len(s)
    num = 0;
    mul = 1;
    sum = 0;
      
    # If the length of given
    # BCD is not divisible by 4   
    for i in range(l % 4 - 1,
                   -1, -1):   
        sum += (ord(s[i]) -
                ord('0')) * mul;
        mul *= 2;
     
    num = sum;
    sum = 0;
    mul = pow(2, 3);
    ctr = 0;
     
    for i in range(l % 4, l):   
        ctr += 1;
        sum += (ord(s[i]) -
                ord('0')) * mul;
        mul //= 2;
        if (ctr == 4):
            num = num * 10 + sum;
            sum = 0;
            mul = pow(2, 3);
            ctr = 0;       
  
    # Convert decimal to binary
    ans = "";
    while (num > 0):
        ans += (chr((num % 2) +
                ord('0')));
        num //= 2;
    ans = ans[:: -1]
     
    return ans;
 
# Driver code
if __name__ == "__main__":
     
    s = "1001000";
  
    # Function Call
    print(bcdToBinary(s));
 
# This code is contributed by rutvik_56

C#

// C# code to convert BCD to its
// equivalent Binary
using System;
using System.Text;
public class GFG{
      
// Function to convert BCD to Decimal
static String bcdToBinary(String s)
{
    int l = s.Length;
    int num = 0;
    int mul = 1;
    int sum = 0;
      
    // If the length of given BCD is not
    // divisible by 4
    for(int i = l % 4 - 1; i >= 0; i--)
    {
        sum += (s[i] - '0') * mul;
        mul *= 2;
    }
    num = sum;
    sum = 0;
    mul = (int)Math.Pow(2, 3);
    int ctr = 0;
    for(int i = l % 4; i < l; i++)
    {
        ctr++;
        sum += (s[i] - '0') * mul;
        mul /= 2;
        if (ctr == 4)
        {
            num = num * 10 + sum;
            sum = 0;
            mul = (int)Math.Pow(2, 3);
            ctr = 0;
        }
    }
      
    // Convert decimal to binary
    String ans = "";
    while (num > 0)
    {
        ans += (char)(num % 2 + '0');
        num /= 2;
    }
    StringBuilder ans1 = new StringBuilder();
  
    // Append a string into StringBuilder input1
    ans1.Append(ans);
  
    // Reverse StringBuilder input1
    ans = reverse(ans1.ToString());
    return ans;
}
static String reverse(String input)
{
        char[] a = input.ToCharArray();
        int l, r = a.Length - 1;
        for (l = 0; l < r; l++, r--)
        {
            char temp = a[l];
            a[l] = a[r];
            a[r] = temp;
        }
        return String.Join("",a);
    }
   
// Driver code
public static void Main(String[] args)
{
    String s = "1001000";
      
    // Function call
    Console.WriteLine(bcdToBinary(s));
}
}
 
// This code is contributed by shikhasingrajput

输出：

110000

时间复杂度： O(N) ，其中N表示提供的BCD字符串的长度
辅助空间： O(1)