程序添加两个二进制字符串

给定两个二进制字符串，返回它们的总和(也是一个二进制字符串)。

例子：

Input:  a = "11", b = "1"
Output: "100" 

我们强烈建议您最小化浏览器，然后先尝试一下
这个想法是从两个字符串的最后一个字符开始，然后一个个地计算数字总和。如果总和大于1，则存储进位为下一位数字。

C++

// C++ program to add two binary strings
#include
using namespace std;
  
// This function adds two binary strings and return
// result as a third string
string addBinary(string a, string b)
{
    string result = ""; // Initialize result
    int s = 0;          // Initialize digit sum
  
    // Traverse both strings starting from last
    // characters
    int i = a.size() - 1, j = b.size() - 1;
    while (i >= 0 || j >= 0 || s == 1)
    {
        // Comput sum of last digits and carry
        s += ((i >= 0)? a[i] - '0': 0);
        s += ((j >= 0)? b[j] - '0': 0);
  
        // If current digit sum is 1 or 3, add 1 to result
        result = char(s % 2 + '0') + result;
  
        // Compute carry
        s /= 2;
  
        // Move to next digits
        i--; j--;
    }
    return result;
}
  
// Driver program
int main()
{
    string a = "1101", b="100";
    cout << addBinary(a, b) << endl;
    return 0;
}

Java

// java program to add 
// two binary strings
  
public class GFG {
  
    // This function adds two 
    // binary strings and return
    // result as a third string
    static String addBinary(String a, String b)
    {
          
        // Initialize result
        String result = ""; 
          
        // Initialize digit sum
        int s = 0;         
  
        // Traverse both strings starting 
        // from last characters
        int i = a.length() - 1, j = b.length() - 1;
        while (i >= 0 || j >= 0 || s == 1)
        {
              
            // Comput sum of last 
            // digits and carry
            s += ((i >= 0)? a.charAt(i) - '0': 0);
            s += ((j >= 0)? b.charAt(j) - '0': 0);
  
            // If current digit sum is 
            // 1 or 3, add 1 to result
            result = (char)(s % 2 + '0') + result;
  
            // Compute carry
            s /= 2;
  
            // Move to next digits
            i--; j--;
        }
          
    return result;
    }
  
    //Driver code
    public static void main(String args[])
    {
        String a = "1101", b="100";
          
        System.out.print(addBinary(a, b));
    }
}
  
// This code is contributed by Sam007.

Python3

# Python Solution for above problem:
  
# This function adds two binary 
# strings return the resulting string
def add_binary_nums(x, y):
        max_len = max(len(x), len(y))
  
        x = x.zfill(max_len)
        y = y.zfill(max_len)
          
        # initialize the result
        result = ''
          
        # initialize the carry
        carry = 0
  
        # Traverse the string
        for i in range(max_len - 1, -1, -1):
            r = carry
            r += 1 if x[i] == '1' else 0
            r += 1 if y[i] == '1' else 0
            result = ('1' if r % 2 == 1 else '0') + result
            carry = 0 if r < 2 else 1     # Compute the carry.
          
        if carry !=0 : result = '1' + result
  
        return result.zfill(max_len)
  
# Driver code 
print(add_binary_nums('1101', '100'))
  
# This code is contributed 
# by Anand Khatri

C#

// C# program to add 
// two binary strings
using System;
  
class GFG {
      
    // This function adds two 
    // binary strings and return
    // result as a third string
    static string addBinary(string a, 
                            string b)
    {
          
        // Initialize result
        string result = ""; 
          
        // Initialize digit sum
        int s = 0;         
  
        // Traverse both strings starting 
        // from last characters
        int i = a.Length - 1, j = b.Length - 1;
        while (i >= 0 || j >= 0 || s == 1)
        {
              
            // Comput sum of last 
            // digits and carry
            s += ((i >= 0)? a[i] - '0': 0);
            s += ((j >= 0)? b[j] - '0': 0);
  
            // If current digit sum is 
            // 1 or 3, add 1 to result
            result = (char)(s % 2 + '0') + result;
  
            // Compute carry
            s /= 2;
  
            // Move to next digits
            i--; j--;
        }
    return result;
    }
      
// Driver Code    
public static void Main()
{
    string a = "1101", b="100";
    Console.Write( addBinary(a, b));
}
}
  
// This code is contributed by Sam007

PHP

= 0 || $j >= 0 || $s == 1)
    {
        // Comput sum of last digits and carry
        $s += (($i >= 0)? ord($a[$i]) - 
                          ord('0'): 0);
        $s += (($j >= 0)? ord($b[$j]) - 
                          ord('0'): 0);
  
        // If current digit sum is 1 or 3, 
        // add 1 to result
        $result = chr($s % 2 + ord('0')) . $result;
  
        // Compute carry
        $s = (int)($s / 2);
  
        // Move to next digits
        $i--; $j--;
    }
    return $result;
}
  
// Driver Code
$a = "1101";
$b = "100";
echo addBinary($a, $b);
  
// This code is contributed by mits
?>

输出：

10001