[leetcode]306. Additive Number

题目链接：https://leetcode.com/problems/additive-number/#/description

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 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.

思路：

se a helper function to add two strings.

Choose first two number then recursively check.

Note that the length of first two numbers can't be longer than half of the initial string, so the two loops in the first function will end when i>num.size()/2 and j>(num.size()-i)/2, this will actually save a lot of time.

Update the case of heading 0s
e.g. "100010" should return false

class Solution {public:        bool isAdditiveNumber(string num) {            for(int i=1; i<=num.size()/2; i++){                for(int j=1; j<=(num.size()-i)/2; j++){                    if(check(num.substr(0,i), num.substr(i,j), num.substr(i+j))) return true;                }            }            return false;        }        bool check(string num1, string num2, string num){            if(num1.size()>1 && num1[0]=='0' || num2.size()>1 && num2[0]=='0') return false;            string sum=add(num1, num2);            if(num==sum) return true;            if(num.size()<=sum.size() || sum.compare(num.substr(0,sum.size()))!=0) return false;            else return check(num2, sum, num.substr(sum.size()));        }         string add(string n, string m){            string res;            int i=n.size()-1, j=m.size()-1, carry=0;            while(i>=0 || j>=0){                int sum=carry+(i>=0 ? (n[i--]-'0') : 0) + (j>=0?  (m[j--]-'0') : 0);                res.push_back(sum%10+'0');                carry=sum/10;            }            if(carry) res.push_back(carry+'0');            reverse(res.begin(), res.end());            return res;        }    };