842. 将数组拆分成斐波那契序列

//确定前三个数
class Solution 
{
public:
    vector<int> splitIntoFibonacci(string num) 
    {
        vector<int> res;
        int n = num.size();
        for(int i = 0;i < n - 2;i++)
        {
            if(i != 0 && num[0] == '0') continue;//"0123"
            for(int j = i + 1;j < n - 1;j ++)
            {
                if(j != i+1 && num[i+1] =='0') continue;//"1023"->"1"和"02"直接退出
                if(i > 9) break; //stol最多只读10位
                long num1 = stol(num.substr(0,i+1));//stol最多只读10位，即 i+1 <= 10 ——> i <= 9
                long num2 = stol(num.substr(i+1, j-i));//stol最多只读10位，即 j-i <= 10 ——> j <= 10+i;
                long num3 = (num1+num2);
                if(num1 > INT_MAX || num2 > INT_MAX || num3 > INT_MAX) break; //0 <= F[i] <= 2^31 - 1
                if( to_string(num3) == num.substr(j+1, to_string(num3).size()))
                {
                    vector<int> temp;
                    temp.push_back(num1);
                    temp.push_back(num2);
                    temp.push_back(num3);
                    string all =  num.substr(0, j+1+to_string(num3).size());
                    while(all.size() < num.size())
                    {
                        num1 = num2;
                        num2 = num3;
                        num3 = num1+num2;
                        if(num1 > INT_MAX || num2 > INT_MAX || num3 > INT_MAX) break; //0 <= F[i] <= 2^31 - 1
                        all+= to_string(num3);
                        temp.push_back(num3);
                    }
                    if(all == num) return temp;
                }
            }
        }
        return {};
    }
};