[LeetCode] Additive Number

Af first I read the title as "Addictive Number". Anyway, this problem can be solved elegantly using recursion. This post shares a nice recursive C++ code with handling of the follow-up by implementing string addition function.

The code is rewritten as follows.

 class Solution {
 public:
     bool isAdditiveNumber(string num) {
         int n = num.size();
         for (int i = ; i <= n/; i++)
             for (int j = ; j <= (n-i)/; j++)
                 if (additive(num.substr(, i), num.substr(i, j), num.substr(i + j)))
                     return true;
         return false;
     }
 private:
     bool additive(string a, string b, string c) {
         string s = add(a, b);
         if (s == c) return true;
         if (c.size() <= s.size() || s.compare(c.substr(, s.size())))
             return false;
         return additive(b, s, c.substr(s.size()));
     }
     string add(string& a, string& b) {
         string s;
         int i = a.size() - , j = b.size() - , c = ;
         while (i >=  || j >= ) {
             int d = (i >=  ? (a[i--] - '') : ) + (j >=  ? (b[j--] - '') : ) + c;
             s += (d %  + '');
             c = d / ;
         }
         if (c) s += ('' + c);
         reverse(s.begin(), s.end());
         return s;
     }
 };

This problem can also be solved iteratively, like peisi's code, which is rewritten in C++ below.

 class Solution {
 public:
     bool isAdditiveNumber(string num) {
         int n = num.length();
         for (int i = ; i <= n/; i++)
             for (int j = ; max(i, j) <= n - i - j; j++)
                 if (additive(i, j, num)) return true;
         return false;
     }
 private:
     bool additive(int i, int j, string& num) {
         if (num[i] == '' && j > ) return false;
         string a = num.substr(, i);
         string b = num.substr(i, j);
         for (int k = i + j; k < num.length(); k += b.length()) {
             b.swap(a);
             b = add(a, b);
             string tail = num.substr(k);
             if (b.compare(tail.substr(, b.size()))) return false;
         }
         return true;
     }
     string add(string& a, string& b) {
         string s;
         int i = a.size() - , j = b.size() - , c = ;
         while (i >=  || j >= ) {
             int d = (i >=  ? (a[i--] - '') : ) + (j >=  ? (b[j--] - '') : ) + c;
             s += (d %  + '');
             c = d / ;
         }
         if (c) s += ('' + c);
         reverse(s.begin(), s.end());
         return s;
     }
 };

If you are a Pythoner, you may also love Stefan's code, which is pasted below.

 def isAdditiveNumber(self, num):
     n = len(num)
     for i, j in itertools.combinations(range(1, n), 2):
         a, b = num[:i], num[i:j]
         if b != str(int(b)):
             continue
         while j < n:
             c = str(int(a) + int(b))
             if not num.startswith(c, j):
                 break
             j += len(c)
             a, b = b, c
         if j == n:
             return True
     return False