Leetcode no. 306

306. 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?

public class Solution {    public boolean isAdditiveNumber(String num) {        for (int i = 1; i <= (num.length()>>1); i++) {            for (int j = 1; j <= (num.length()>>1); j++) {                String first= num.substring(0,i);                if (2*i+j>num.length() || 2*j+i>num.length()) break;                String second= num.substring(i,i+j);                String rest= num.substring(i+j);                if (checkAdditive(first, second, rest)==true) return true;            }        }        return false;    }    private boolean checkAdditive(String s1, String s2, String s){        if ((s1.length()>1 && s1.charAt(0)=='0') || (s2.length()>1 && s2.charAt(0)=='0'))            return false;        String add= addUp(s1,s2);        if (add.equals(s)) return true;        if (s.length()<add.length()) return false;        if (!s.substring(0, add.length()).equals(add)) return false;        return checkAdditive(s2, add, s.substring(add.length()));    }    private String addUp(String s1, String s2){        StringBuilder sb= new StringBuilder();        int i= s1.length()-1, j= s2.length()-1, flag=0;        while (i>=0 || j>=0 || flag>0){            int add= flag+ ((i>=0)? s1.charAt(i--)-'0':0)+((j>=0)? s2.charAt(j--)-'0':0);            sb.insert(0, add%10);            flag=add/10;        }        return sb.toString();    }}