Long Number Operation Sep 25, 2017
Multiply
//`num1[i] * num2[j]` will be placed at indices `[i + j`, `i + j + 1]`
public String multiply(String num1, String num2) {
int m = num1.length(), n = num2.length();
int[] pos = new int[m + n];
for(int i = m - 1; i >= 0; i--) {
for(int j = n - 1; j >= 0; j--) {
int mul = (num1.charAt(i) - '0') * (num2.charAt(j) - '0');
int p1 = i + j, p2 = i + j + 1;
int sum = mul + pos[p2];
pos[p1] += sum / 10;
pos[p2] = (sum) % 10;
}
}
StringBuilder sb = new StringBuilder();
for(int p : pos) if(!(sb.length() == 0 && p == 0)) sb.append(p);
return sb.length() == 0 ? "0" : sb.toString();
}
Add
//Core operation
int carry = 0;
int digit = 0;
for(int i=0 ; i<n1.length() ; i++)
{
int digitResult = oneDigitAdd(n1.charAt(i), n2.charAt(i)) + carry; //key part
carry = digitResult/10;
digit = digitResult%10;
result.append(digit);
}
result.append(carry); // append the final carry
Back;