Leetcode January 10 daily question 306 Additive number

306. Additive number

Thought analysis ：
Enumerate the first and second numbers , The third number is calculated by high-precision addition , And then compare them

AC Code ：

// C++
class Solution {
    
public:
    //  High precision addition board 
    string add(string x, string y) {
    
        vector<int> A, B, C;![ Please add a picture description ](https://img-blog.csdnimg.cn/d5eaa9b0fe2e4870ba30d7c7c01419a6.png?x-oss-process=image/watermark,type_d3F5LXplbmhlaQ,shadow_50,text_Q1NETiBA5Zev5oiR5oOz5oOz,size_20,color_FFFFFF,t_70,g_se,x_16)

        //  First the  x,y  Reverse deposit  vector
        for(int i = x.size() - 1; i >= 0; i--) A.push_back(x[i] - '0');
        for(int i = y.size() - 1; i >= 0; i--) B.push_back(y[i] - '0');
        //  Analog addition begins 
        for(int i = 0, t = 0; i < A.size() || i < B.size() || t; i++) {
    
            if(i < A.size()) t += A[i];
            if(i < B.size()) t += B[i];
            C.push_back(t % 10);
            t /= 10;
        }
        string z;
        for(int i = C.size() - 1; i >=0; i--) z += to_string(C[i]);
        return z;
    }

    bool isAdditiveNumber(string num) {
    
        for(int i = 0; i < num.size(); i++)
            for(int j = i + 1; j + 1 < num.size(); j++) {
    
                //  First number  a+1  To  b
                //  The second number  b+1  To  c
                //  The third number  c+1  To  ... 
                //  therefore a Initial value of from  -1  Start  
                int a = -1, b = i, c = j;
                while(true) {
    
                    if((b - a > 1 && num[a + 1] == '0') || (c - b > 1 && num[b + 1] == '0')) break;    //  A leading 0  illegal 
                    auto x = num.substr(a + 1, b - a), y = num.substr(b + 1, c - b);
                    auto z = add(x, y);
                    if(num.substr(c + 1, z.size()) != z) break;
                    a = b, b = c, c += z.size();
                    if(c + 1 == num.size()) return true;
                }
            }   
        return false;
    }
};