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?

Credits:

Special thanks to @jeantimex for adding this problem and creating all test cases.

Brute Force,要注意的是题目中的提示，以0开头的两位数或多位数是不合法的，还有就是循环的起止条件，两数相加等于第三数，第三个数的长度一定是大于等于前两个数之中长度最长的数的，代码：

public boolean isAdditiveNumber(String num) {      int len=num.length();    for(int i=1;i<=len/2;i++){        for(int j=1;Math.max(i,j)<=len-i-j;j++){            if(isAddi(num,i,j)) return true;        }    }    return false;}public boolean isAddi(String num,int i,int j){    if(num.charAt(0)=='0'&&i>1) return false;    if(num.charAt(i)=='0'&&j>1) return false;    Long n1=Long.parseLong(num.substring(0,i));    Long n2=Long.parseLong(num.substring(i,i+j));    String sum="";    for(int k=i+j;k!=num.length();k+=sum.length()){        n2=n1+n2;        n1=n2-n1;        sum=n2.toString();        if(!num.startsWith(sum,k)) return false;    }    return true;}