Dadas dos strings binarias, devuelva su suma (también una string binaria).
Ejemplo:
C
// C++ program to add two binary strings
#include <bits/stdc++.h>
using namespace std;
// This function adds two binary strings and return
// result as a third string
string addBinary(string A, string B)
{
// If the length of string A is greater than the length
// of B then just swap the string by calling the
// same function and make sure to return the function
// otherwise recursion will occur which leads to
// calling the same function twice
if (A.length() > B.length())
return addBinary(B, A);
// Calculating the difference between the length of the
// two strings.
int diff = B.length() - A.length();
// Initialise the padding string which is used to store
// zeroes that should be added as prefix to the string
// which has length smaller than the other string.
string padding;
for (int i = 0; i < diff; i++)
padding.push_back('0');
A = padding + A;
string res;
char carry = '0';
for (int i = A.length() - 1; i >= 0; i--) {
// This if condition solves 110 111 possible cases
if (A[i] == '1' && B[i] == '1') {
if (carry == '1')
res.push_back('1'), carry = '1';
else
res.push_back('0'), carry = '1';
}
// This if condition solves 000 001 possible cases
else if (A[i] == '0' && B[i] == '0') {
if (carry == '1')
res.push_back('1'), carry = '0';
else
res.push_back('0'), carry = '0';
}
// This if condition solves 100 101 010 011 possible
// cases
else if (A[i] != B[i]) {
if (carry == '1')
res.push_back('0'), carry = '1';
else
res.push_back('1'), carry = '0';
}
}
// If at the end their is carry then just add it to the
// result
if (carry == '1')
res.push_back(carry);
// reverse the result
reverse(res.begin(), res.end());
// To remove leading zeroes
int index = 0;
while (index + 1 < res.length() && res[index] == '0')
index++;
return (res.substr(index));
}
// 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)
{
//If the inputs are 0
if(a.charAt(0) == '0' && b.charAt(0) == '0'){
return "0";
}
// Initialize result
StringBuilder result = new StringBuilder("");
// 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.append((char)(s % 2 + '0'));
// Compute carry
s /= 2;
// Move to next digits
i--; j--;
}
// Remove leading zeros, if any
int start = result.length()-1;
while(start >=0 && result.charAt(start) == '0') {
start--;
}
if(start != result.length()-1) {
result.delete(start+1,result.length());
}
return result.reverse().toString();
}
//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
<?php
// PHP program to add two binary strings
// This function adds two binary strings
// and return result as a third string
function addBinary($a, $b)
{
$result = ""; // Initialize result
$s = 0; // Initialize digit sum
// Traverse both strings starting
// from last characters
$i = strlen($a) - 1;
$j = strlen($b) - 1;
while ($i >= 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
?>
Javascript
<script>
// Javascript program to add
// two binary strings
// This function adds two
// binary strings and return
// result as a third string
function addBinary(a, b)
{
// Initialize result
var result = "";
// Initialize digit sum
var s = 0;
// Traverse both strings starting
// from last characters
var 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).charCodeAt(0) -
'0'.charCodeAt(0): 0);
s += ((j >= 0)? b.charAt(j).charCodeAt(0) -
'0'.charCodeAt(0): 0);
// If current digit sum is
// 1 or 3, add 1 to result
result = String.fromCharCode(parseInt(s % 2) +
'0'.charCodeAt(0)) + result;
// Compute carry
s = parseInt(s/2);
// Move to next digits
i--; j--;
}
return result;
}
//Driver code
var a = "1101", b="100";
document.write(addBinary(a, b));
// This code is contributed by Amit Katiyar
</script>
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA