Solution：digit_sum = prev_carry + a[i] + b[i], calculating backwards
Time Complexity: O(max_length(a, b))
Space Complexity: O(max_length(a, b)) for StringBuilder
Solution Code：
class Solution {
public String addBinary(String a, String b) {
StringBuilder result = new StringBuilder();
int i = a.length() - 1, j = b.length() - 1, carry = 0;
while(i >= 0 || j >= 0) {
int sum = carry;
if(i >= 0) sum += a.charAt(i--) - '0';
if(j >= 0) sum += b.charAt(j--) - '0';
result.append(sum % 2);
carry = sum / 2;
}
if(carry != 0) result.append(carry);
return result.reverse().toString();
}
}
