LeetCode-Additive Number

Additive number is a string whose digits can form additive sequence.

A valid additive sequence should contain at least three numbers. Except for the first two numbers, each subsequent number in the sequence must be the sum of the preceding two.

For example:
"112358" is an additive number because the digits can form an additive sequence: 1, 1, 2, 3, 5, 8.

1 + 1 = 2, 1 + 2 = 3, 2 + 3 = 5, 3 + 5 = 8

"199100199" is also an additive number, the additive sequence is: 1, 99, 100, 199.

1 + 99 = 100, 99 + 100 = 199

Note: Numbers in the additive sequence cannot have leading zeros, so sequence 1, 2, 03 or 1, 02, 3 is invalid.

Given a string containing only digits '0'-'9', write a function to determine if it's an additive number.

Follow up:
How would you handle overflow for very large input integers?

Analysis:

All varieties come from the first two numbers of the input string. If that is determined, all the followings are just confirming the addition sequence. 

Solution:

public class Solution {
   public boolean isAdditiveNumber(String num) {
        if (num.length() < 3)
            return false;

        for (int len1 = 1; len1 <= num.length() / 2; len1++) {
            int maxLen2 = Math.min(num.length() - 2 * len1, (num.length() - len1) / 2);
            for (int len2 = 1; len2 <= maxLen2; len2++)
                if (isAddNumRecur(num, 0, len1, len1 + len2)) {
                    return true;
                }
        }
        return false;
    }

    // Is num an additive number, if we set the first and second number as
    // num(p1,p2-1) and num(p2,p3-1).
    public boolean isAddNumRecur(String num, int p1, int p2, int p3) {
        if (p3 == num.length())
            return true;

        
        // Get the two numbers to start with and get their sum.
        long num1 = Long.parseLong(num.substring(p1, p2));
        long num2 = Long.parseLong(num.substring(p2, p3));
        // the two numbers cannot start with '0' if they are not 0
        if ((num1!=0 && num.charAt(p1) == '0') || (num2!=0 && num.charAt(p2) == '0'))
            return false;
        
        long num3 = num1 + num2;
        String num3Str = Long.toString(num3);
        if (num3Str.length() > num.length() - p3)
            return false;

        // If starting from p3, we can find a number that equals to num(p1,p2-1)
        // + num(p2,p3-1)
        for (int i = 0; i < num3Str.length(); i++)
            if (num.charAt(p3 + i) != num3Str.charAt(i)) {
                return false;
            }

        return isAddNumRecur(num, p2, p3, p3 + num3Str.length());
    }
}