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// 306. Additive Number
#include "leetcode.h"
/*
* an additive number is a string whose digits can form an additive sequence. a
* valid additive seuqnece should contain at least three numbers. except for the
* firs ttwo numbers each subsequent number in the sequence must be the sum of
* the preceding two. given a string containing only digits, return 'true' if it
* is an additive number, or 'false' otherwise. note, numbers in the additive
* sequence cannot have leading zeros, so sequence, 1,2,03, or 1,02,3 is invalid
*/
class Solution {
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;
}
bool check(string n1, string n2, string n) {
if (n1.size() > 1 && n1[0] == '0' || n2.size() > 1 && n2[0] == '0')
return false;
string sum = add(n1, n2);
if (n == sum)
return true;
if (n.size() <= sum.size() || !sum.compare(n.substr(0, sum.size())))
return false;
else
return check(n1, sum, n.substr(sum.size()));
}
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;
}
};
int main() {
Solution obj;
printf("%d\n", obj.isAdditiveNumber("112358")); // expect: 1
printf("%d\n", obj.isAdditiveNumber("199100199")); // expect: 1
};
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