给定两个相等长度的二进制字符串A和B ,任务是打印一个字符串,它是二进制字符串A和B的异或。
例子:
Input: A = “0001”, B = “0010”
Output: 0011
Input: A = “1010”, B = “0101”
Output: 1111
方法:我们的想法是遍历由字符字符串的字符都和如果字符不匹配然后加“1”在回答字符串的字符,否则加“0”的答案字符串生成XOR字符串。
下面是上述方法的实现:
C++
// C++ Implementation to find the
// XOR of the two Binary Strings
#include
using namespace std;
// Function to find the
// XOR of the two Binary Strings
string xoring(string a, string b, int n){
string ans = "";
// Loop to iterate over the
// Binary Strings
for (int i = 0; i < n; i++)
{
// If the Character matches
if (a[i] == b[i])
ans += "0";
else
ans += "1";
}
return ans;
}
// Driver Code
int main()
{
string a = "1010";
string b = "1101";
int n = a.length();
string c = xoring(a, b, n);
cout << c << endl;
}
// This code is contributed by Surendra_Gangwar
Java
// Java Implementation to find the
// XOR of the two Binary Strings
import java.io.*;
class GFG {
// Function to find the
// XOR of the two Binary Strings
static String xoring(String a, String b, int n){
String ans = "";
// Loop to iterate over the
// Binary Strings
for (int i = 0; i < n; i++)
{
// If the Character matches
if (a.charAt(i) == b.charAt(i))
ans += "0";
else
ans += "1";
}
return ans;
}
// Driver Code
public static void main (String[] args)
{
String a = "1010";
String b = "1101";
int n = a.length();
String c = xoring(a, b, n);
System.out.println(c);
}
}
// This code is contributed by shubhamsingh10
Python3
# Python Implementation to find the
# XOR of the two Binary Strings
# Function to find the
# XOR of the two Binary Strings
def xor(a, b, n):
ans = ""
# Loop to iterate over the
# Binary Strings
for i in range(n):
# If the Character matches
if (a[i] == b[i]):
ans += "0"
else:
ans += "1"
return ans
# Driver Code
if __name__ == "__main__":
a = "1010"
b = "1101"
n = len(a)
c = xor(a, b, n)
print(c)
C#
// C# Implementation to find the
// XOR of the two Binary Strings
using System;
class GFG{
// Function to find the
// XOR of the two Binary Strings
static string xoring(string a, string b, int n){
string ans = "";
// Loop to iterate over the
// Binary Strings
for (int i = 0; i < n; i++)
{
// If the Character matches
if (a[i] == b[i])
ans += "0";
else
ans += "1";
}
return ans;
}
// Driver Code
static public void Main ()
{
string a = "1010";
string b = "1101";
int n = a.Length;
string c = xoring(a, b, n);
Console.WriteLine(c);
}
}
// This code is contributed by shubhamsingh10
Javascript
输出:
0111