Leetcode196: Additive Number

dditive 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.

class Solution {public:

//两个字符串相加    string strAdd(string &v1, string &v2){        string res = "";        int carry = 0;        int len1 = v1.size(), len2 = v2.size();            for(int i = len1-1, j = len2-1; i >= 0 || j >= 0; i--, j--){            int n1 = (i >= 0) ? v1[i]-'0' : 0;            int n2 = (j >= 0) ? v2[j]-'0' : 0;            res = to_string((n1 + n2 + carry) % 10) + res;            carry = (n1 + n2 + carry) > 9;        }            return (carry) ? "1" + res : res;    }//回溯法，比较v1+v2的值是否在num之后出现，如果出现则继续往后比较v2+sumstr的值。。。    bool isAdditive(string &num, string v1, string v2, int idx, int n){        string sumStr = strAdd(v1, v2);        if(num.compare(idx, sumStr.size(), sumStr) == 0)            return (idx + sumStr.size() >= n || isAdditive(num, v2, sumStr, idx+sumStr.size(), n));        else            return false;    }//从首位开始尝试，一旦found=true则存在    bool isAdditiveNumber(string num) {        int n = num.size();        bool found = 0;            for(int len1 = 1; len1 <= (n/2); len1++){            for(int len2 = 1; len2 <= (n/2); len2++){                found = isAdditive(num, num.substr(0, len1), num.substr(len1, len2), len1+len2, n);                if(found)                       return true;            }        }        return false;    }};