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 sequence1, 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.

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class Solution {    bool isAdd(float nums1,float nums2,string num)    {        if(num==""||num.size()>0&&num[0]=='0') return false;//进入这个循环说明需要比较，要求出第三个数，                                                          // 因此如果为空或者第一个数为0returnfalse        float sum=nums1+nums2;        float total =stol(num);                     //求总的值，利于退出        int len=log10(sum)+1;                       //求出合的位数。        if(sum==total) return true;                 //如果可以，一步退出。        if(total<sum || num.size()<len) return false;   //如果和大于剩余总的，可以return false.        float nums3=stol(num.substr(0,len));        //求出第三个数判断，长度由1和2决定        if(nums3!=sum) return false;                //不想等应该直接return false.        else        return isAdd(nums2,nums3,num.substr(len));//相等则进入下一层循环，nums1=nums2,nums2=nums3    }public:    bool isAdditiveNumber(string num) {        int len=num.size();        for(int i=1;i<=len/2;++i)        {            if(num[0]=='0'&&i>1) break; //0可以单独.出现，因此i>1时候需要break;            int nums1=stol(num.substr(0,i));            for(int j=1;j<=len/2;++j)            {                if(num[i]=='0'&&j>1)//0可以单独.出现，因此j>1时候需要break;                break;                int nums2=stol(num.substr(i,j));                if(isAdd(nums1,nums2,num.substr(i+j))==true)//外层控制第一层的两个数                return true;            }        }        return false;    }};