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;
+
+ }
+};