给定两个位序列作为字符串，编写一个函数以返回两个序列的加法。位字符串也可以具有不同的长度。例如，如果字符串1为“ 1100011”，第二个字符串2为“ 10”，则该函数应返回“ 1100101”。
我们强烈建议您单击此处并进行实践，然后再继续解决方案。
由于两个字符串的大小可能不同，因此我们首先通过添加前导0来使较小的字符串的大小等于较大的字符串的大小。在使大小相同之后，我们从最右边的位到最左边的位一一加法。在每次迭代中，我们需要对3位求和：2个给定字符串的2位并进位。如果全部设置了3个位或设置了其中3个位，则总和为1。因此，我们可以对所有位进行XOR运算以找到总和位。如何找到进位–如果两位中的任意一位被置位，进位将为1。因此，我们可以通过对所有对进行“或”运算来找到进位。以下是逐步算法。
1.在较小的字符串开头加上0，以使其大小相等。
2.执行位加法
…..用于添加3位a，b，c的布尔表达式
…..Sum = a XOR b XOR c
…..Carry =(a AND b)或(b AND c)OR(c AND a)
以下是上述算法的实现。
C++
#include
using namespace std;
//adds the two-bit strings and return the result
string addBitStrings( string first, string second );
// Helper method: given two unequal sized bit strings, converts them to
// same length by adding leading 0s in the smaller string. Returns the
// the new length
int makeEqualLength(string &str1, string &str2)
{
int len1 = str1.size();
int len2 = str2.size();
if (len1 < len2)
{
for (int i = 0 ; i < len2 - len1 ; i++)
str1 = '0' + str1;
return len2;
}
else if (len1 > len2)
{
for (int i = 0 ; i < len1 - len2 ; i++)
str2 = '0' + str2;
}
return len1; // If len1 >= len2
}
// The main function that adds two-bit sequences and returns the addition
string addBitStrings( string first, string second )
{
string result; // To store the sum bits
// make the lengths same before adding
int length = makeEqualLength(first, second);
int carry = 0; // Initialize carry
// Add all bits one by one
for (int i = length-1 ; i >= 0 ; i--)
{
int firstBit = first.at(i) - '0';
int secondBit = second.at(i) - '0';
// boolean expression for sum of 3 bits
int sum = (firstBit ^ secondBit ^ carry)+'0';
result = (char)sum + result;
// boolean expression for 3-bit addition
carry = (firstBit & secondBit) | (secondBit & carry) | (firstBit & carry);
}
// if overflow, then add a leading 1
if (carry)
result = '1' + result;
return result;
}
// Driver program to test above functions
int main()
{
string str1 = "1100011";
string str2 = "10";
cout << "Sum is " << addBitStrings(str1, str2);
return 0;
} Java
// Java implementation of above algorithm
class GFG
{
// Helper method: given two unequal sized bit strings,
// converts them to same length by adding leading 0s
// in the smaller string. Returns the the new length
// Using StringBuilder as Java only uses call by value
static int makeEqualLength(StringBuilder str1,
StringBuilder str2)
{
int len1 = str1.length();
int len2 = str2.length();
if (len1 < len2)
{
for (int i = 0; i < len2 - len1; i++)
str1.insert(0, '0');
return len2;
}
else if (len1 > len2)
{
for (int i = 0; i < len1 - len2; i++)
str2.insert(0, '0');
}
return len1; // If len1 >= len2
}
// The main function that adds two-bit sequences
// and returns the addition
static String addBitStrings(StringBuilder str1,
StringBuilder str2)
{
String result = ""; // To store the sum bits
// make the lengths same before adding
int length = makeEqualLength(str1, str2);
// Convert StringBuilder to Strings
String first = str1.toString();
String second = str2.toString();
int carry = 0; // Initialize carry
// Add all bits one by one
for (int i = length - 1; i >= 0; i--)
{
int firstBit = first.charAt(i) - '0';
int secondBit = second.charAt(i) - '0';
// boolean expression for sum of 3 bits
int sum = (firstBit ^ secondBit ^ carry) + '0';
result = String.valueOf((char) sum) + result;
// boolean expression for 3-bit addition
carry = (firstBit & secondBit) |
(secondBit & carry) |
(firstBit & carry);
}
// if overflow, then add a leading 1
if (carry == 1)
result = "1" + result;
return result;
}
// Driver Code
public static void main(String[] args)
{
String str1 = "1100011";
String str2 = "10";
System.out.println("Sum is " +
addBitStrings(new StringBuilder(str1),
new StringBuilder(str2)));
}
}
// This code is contributed by Vivek Kumar SinghPython3
# Python3 program for above approach
# adds the two-bit strings and return the result
# Helper method: given two unequal sized bit strings,
# converts them to same length by adding leading 0s
# in the smaller string. Returns the the new length
def makeEqualLength(str1, str2):
len1 = len(str1) # Length of string 1
len2 = len(str2) # length of string 2
if len1 < len2:
str1 = (len2 - len1) * '0' + str1
len1 = len2
elif len2 < len1:
str2 = (len1 - len2) * '0' + str2
len2 = len1
return len1, str1, str2
def addBitStrings( first, second ):
result = '' # To store the sum bits
# make the lengths same before adding
length, first, second = makeEqualLength(first, second)
carry = 0 # initialize carry as 0
# Add all bits one by one
for i in range(length - 1, -1, -1):
firstBit = int(first[i])
secondBit = int(second[i])
# boolean expression for sum of 3 bits
sum = (firstBit ^ secondBit ^ carry) + 48
result = chr(sum) + result
# boolean expression for 3 bits addition
carry = (firstBit & secondBit) | \
(secondBit & carry) | \
(firstBit & carry)
# if overflow, then add a leading 1
if carry == 1:
result = '1' + result
return result
# Driver Code
if __name__ == '__main__':
str1 = '1100011'
str2 = '10'
print('Sum is', addBitStrings(str1, str2))
# This code is contributed by
# chaudhary_19 (Mayank Chaudhary)C#
// C# implementation of above algorithm
using System;
using System.Text;
class GFG
{
// Helper method: given two unequal sized
// bit strings, converts them to same length
// by adding leading 0s in the smaller string.
// Returns the the new length Using StringBuilder
// as Java only uses call by value
static int makeEqualLength(StringBuilder str1,
StringBuilder str2)
{
int len1 = str1.Length;
int len2 = str2.Length;
if (len1 < len2)
{
for (int i = 0; i < len2 - len1; i++)
{
str1.Insert(0, '0');
}
return len2;
}
else if (len1 > len2)
{
for (int i = 0; i < len1 - len2; i++)
{
str2.Insert(0, '0');
}
}
return len1; // If len1 >= len2
}
// The main function that adds two-bit sequences
// and returns the addition
static string addBitStrings(StringBuilder str1,
StringBuilder str2)
{
string result = ""; // To store the sum bits
// make the lengths same before adding
int length = makeEqualLength(str1, str2);
// Convert StringBuilder to Strings
string first = str1.ToString();
string second = str2.ToString();
int carry = 0; // Initialize carry
// Add all bits one by one
for (int i = length - 1; i >= 0; i--)
{
int firstBit = first[i] - '0';
int secondBit = second[i] - '0';
// boolean expression for sum of 3 bits
int sum = (firstBit ^ secondBit ^ carry) + '0';
result = ((char) sum).ToString() + result;
// boolean expression for 3-bit addition
carry = (firstBit & secondBit) |
(secondBit & carry) |
(firstBit & carry);
}
// if overflow, then add a leading 1
if (carry == 1)
{
result = "1" + result;
}
return result;
}
// Driver Code
public static void Main(string[] args)
{
string str1 = "1100011";
string str2 = "10";
Console.WriteLine("Sum is " +
addBitStrings(new StringBuilder(str1),
new StringBuilder(str2)));
}
}
// This code is contributed by kumar65输出：
Sum is 1100101