306. Additive Number ----------M 一开始没有考虑 overflow的问题

 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.

Given a string containing only digits '0'-'9', write a function to determine if it's an additive number.

Note: Numbers in the additive sequence cannot have leading zeros, so sequence 1, 2, 03 or 1, 02, 3 is invalid.

 

Example 1:

Input: "112358"
Output: true
Explanation: 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

Example 2:

Input: "199100199"
Output: true
Explanation: The additive sequence is: 1, 99, 100, 199. 
             1 + 99 = 100, 99 + 100 = 199

 

Constraints:

• num consists only of digits '0'-'9'.
• 1 <= num.length <= 35

Follow up:
How would you handle overflow for very large input integers?

A:

TNND,  凌晨，4，5点钟，真是伤脑子。 再不熬夜了。  

一个SB的错误， 应该时continue, 我因为以前写成了return false.  搞了半个小时

class Solution {
public:
    bool isAdditiveNumber(string num) {
        int n = num.length();
        if( n==0 )
            return true;
        for(int end1 = 0;end1<n-2;end1++){ // end of first number, inclusive            
            if(end1 > 0 && num[0] == '0')
                return false;
            for(int end2 = end1+1; end2+1<n;end2++){
                int len1 = end1+1;
                int len2 = end2 - end1;
                if(len2 > 1 && num[end1+1] == '0')
                    continue;
                if(helper(num.substr(0, len1), num.substr(end1+1,len2), num.substr(end2+1))){
                    return true;
                }
            }
        }
        return false;
    }
private:
    bool helper(string a, string b, string num){
        int carry = 0;
        while(num.length()>0){
            string c = getSum(a,b);
            if(num.find(c) != 0)
                return false;
            num = num.substr(c.length());
            a = b;
            b = c;
        }
        return true;
    }
    string getSum(string a, string b){
        string res = "";
        int carry = 0;
        int na = a.length(), nb = b.length();
        for(int i = 0; i<max(na, nb); i++){
            int c1 = (i<na)? a[na-1-i]-'0':0;
            int c2 = (i<nb)? b[nb-1-i]-'0':0;
            res = to_string( (c1+c2+carry)%10 ) + res;
            carry = (c1 + c2 + carry) / 10;
        }
        if(carry)
            res = "1"+res;
        return res;
    }
};