Additive Number

Additive number is a positive integer 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 represents an integer, write a function to determine if it's an additive number.

Follow up:

How would you handle overflow for very large input integers?

class Solution {public:    bool isAdditiveNumber(string num) {        int len=num.size();        for(int i=1;i<=len/2;i++){            for(int j=1;j<=(len-i)/2;j++){                string str1=num.substr(0,i);                string str2=num.substr(i,j);                if(check2(str1,str2,num.substr(i+j))) return true;            }        }        return false;    }    bool check(string str1,string str2,string num){        string sum=add(str1,str2);        int n3=sum.size();        if(sum==num) return true;        if(n3>=num.size()||sum!=num.substr(0,n3)) return false;        else return check(str2,sum,num.substr(n3));    }    bool check2(string str1,string str2,string num){        string sum=add(str1,str2);        while(true){            if(sum==num) return true;            if(sum.size()>=num.size()||sum!=num.substr(0,sum.size())) return false;            num=num.substr(sum.size());            str1=str2;str2=sum;            sum=add(str1,str2);        }    }    string add(string str1,string str2){        int i1=str1.size()-1;        int i2=str2.size()-1;        int carry=0;        string res;        while(i1>=0||i2>=0){            int sum=(i1>=0?str1[i1]-'0':0)+(i2>=0?str2[i2]-'0':0)+carry;            carry=sum/10;            res.push_back(sum%10+'0');            i1--;i2--;        }        if(carry) res.push_back(carry+'0');        reverse(res.begin(),res.end());        return res;    }};