New Problem "Additive Number"

README.md

@@ -8,6 +8,7 @@ LeetCode
 
 | # | Title | Solution | Difficulty |
 |---| ----- | -------- | ---------- |
+|306|[Additive Number](https://leetcode.com/problems/additive-number/) | [C++](./algorithms/cpp/additiveNumber/AdditiveNumber.cpp)|Medium|
 |304|[Range Sum Query 2D - Immutable](https://leetcode.com/problems/range-sum-query-2d-immutable/) | [C++](./algorithms/cpp/rangeSumQuery2D-Immutable/RangeSumQuery2dImmutable.cpp)|Medium|
 |303|[Range Sum Query - Immutable](https://leetcode.com/problems/range-sum-query-immutable/)  | [C++](./algorithms/cpp/rangeSumQuery-Immutable/rangeSumQuery-Immutable.cpp)|Easy|
 |301|[Remove Invalid Parentheses](https://leetcode.com/problems/remove-invalid-parentheses/) | [C++](./algorithms/cpp/removeInvalidParentheses/RemoveInvalidParentheses.cpp) |Hard|

algorithms/cpp/additiveNumber/AdditiveNumber.cpp

@@ -0,0 +1,95 @@
+// Source : https://leetcode.com/problems/additive-number/
+// Author : Hao Chen
+// Date   : 2015-11-22
+
+/*************************************************************************************** 
+ *
+ * Additive number is a positive integer 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 represents an integer, write a function to determine if it's an 
+ * additive number.
+ * 
+ * Follow up:
+ * How would you handle overflow for very large input integers?
+ * 
+ * Credits:Special thanks to @jeantimex for adding this problem and creating all test 
+ * cases.
+ *               
+ ***************************************************************************************/
+
+
+class Solution {
+public:
+    bool isAdditiveNumber(string num) {
+        int len = num.size();
+        
+        for(int i=1; i<len/2+1; i++) {
+            string n1 = num.substr(0, i);
+            if ( n1.size()>1 && n1[0] == '0') break;
+            for(int j=i+1; j<len; j++) {
+                string n2 = num.substr(i, j-i);
+                if ( n2.size()>1 && n2[0] == '0') break;
+                string n3 = num.substr(j);
+                if (isAdditiveNumberHelper(n1, n2, n3)) return true;
+            }
+        }
+        return false;
+    }
+    
+private:
+    bool isAdditiveNumberHelper(string& n1, string& n2, string& n3){
+        string add = StringAdd(n1, n2);
+        
+        if (add.size() > n3.size()) return false;
+        
+        if (add == n3 ) return true;
+        
+        //split the n3 to 2 parts, and keep going.
+        string cut = n3.substr(0, add.size());
+        if (add == cut) {
+            string rest = n3.substr(add.size());
+            return isAdditiveNumberHelper(n2, add, rest);
+        }
+        return false;
+    }
+
+
+    string StringAdd(string n1, string n2) {
+        
+        if (n1.size() < n2.size()) {
+            string tmp = n1;
+            n1 = n2;
+            n2 = tmp;
+        }
+        
+        int carry=0;
+        string result;
+        for (int i=n1.size()-1, j=n2.size()-1; i>=0; i--, j--) {
+
+            int n = n1[i] - '0' + carry;
+            if ( j >= 0) {
+                n += n2[j] - '0';
+            } 
+            char ch = n % 10 + '0';
+            carry = n/10;
+            result = ch + result;
+        }
+        if (carry>0) result = (char)(carry+'0') + result;
+        return result;
+        
+    }
+};