class BigInt
{
private:
string data;
bool isNegative;
};
BigInt multiplication(BigInt left, BigInt right)
{
BigInt sum;
BigInt result;
sum.da
This is a solution from geeksforgeeks which is very similar to what you are trying to do. I modified it to fit your class there might be an error as I have not compiled it.
BigInt multiplication(BigInt num1, BigInt num2)
{
int n1 = num1.data.size();
int n2 = num2.data.size();
if (n1 == 0 || n2 == 0)
return "0";
// will keep the result number in vector
// in reverse order
vector result(n1 + n2, 0);
// Below two indexes are used to find positions
// in result.
int i_n1 = 0;
int i_n2 = 0;
// Go from right to left in num1
for (int i=n1-1; i>=0; i--)
{
int carry = 0;
int n1 = num1.data[i] - '0';
// To shift position to left after every
// multiplication of a digit in num2
i_n2 = 0;
// Go from right to left in num2
for (int j=n2-1; j>=0; j--)
{
// Take current digit of second number
int n2 = num2[j].data - '0';
// Multiply with current digit of first number
// and add result to previously stored result
// at current position.
int sum = n1*n2 + result[i_n1 + i_n2] + carry;
// Carry for next iteration
carry = sum/10;
// Store result
result[i_n1 + i_n2] = sum % 10;
i_n2++;
}
// store carry in next cell
if (carry > 0)
result[i_n1 + i_n2] += carry;
// To shift position to left after every
// multiplication of a digit in num1.
i_n1++;
}
// ignore '0's from the right
int i = result.size() - 1;
while (i>=0 && result[i] == 0)
i--;
// If all were '0's - means either both or
// one of num1 or num2 were '0'
if (i == -1)
return "0";
// generate the result string
string s = "";
while (i >= 0)
s += std::to_string(result[i--]);
BigInt temp(s, num1.isNegative ^ num2.isNegative);
return temp;
}
Hope this helps.