New Problem "Integer to English Words"

Author: haoel

Diff for: README.md

@@ -13,6 +13,7 @@ LeetCode
 |285|[Inorder Successor in BST](https://leetcode.com/problems/inorder-successor-in-bst/) ♥ | [Java](./algorithms/java/src/inorderSuccessorInBST/inorderSuccessorInBST.java)|Medium|
 |283|[Move Zeroes](https://leetcode.com/problems/move-zeroes/)  | [C++](./algorithms/cpp/moveZeroes/moveZeroes.cpp)|Easy|
 |278|[First Bad Version](https://leetcode.com/problems/first-bad-version/)| [C++](./algorithms/cpp/firstBadVersion/FirstBadVersion.cpp), [Java](./algorithms/java/src/firstBadVersion/firstBadVersion.java)|Easy|
+|273|[Integer to English Words](https://leetcode.com/problems/integer-to-english-words/)| [C++](./algorithms/cpp/integerToEnglishWords/IntegerToEnglishWords.cpp)|Medium|
 |268|[Missing Number](https://leetcode.com/problems/missing-number/)| [C++](./algorithms/cpp/missingNumber/MissingNumber.cpp)|Medium|
 |264|[Ugly Number II](https://leetcode.com/problems/ugly-number/)| [C++](./algorithms/cpp/uglyNumber/UglyNumber.II.cpp)|Medium|
 |263|[Ugly Number](https://leetcode.com/problems/ugly-number/)| [C++](./algorithms/cpp/uglyNumber/UglyNumber.cpp)|Easy|

Diff for: algorithms/cpp/integerToEnglishWords/IntegerToEnglishWords.cpp

@@ -0,0 +1,131 @@
+// Source : https://leetcode.com/problems/integer-to-english-words/
+// Author : Hao Chen
+// Date   : 2015-10-22
+
+/*************************************************************************************** 
+ *
+ * Convert a non-negative integer to its english words representation. Given input is 
+ * guaranteed to be less than 231 - 1.
+ * 
+ * For example,
+ * 
+ * 123 -> "One Hundred Twenty Three"
+ * 12345 -> "Twelve Thousand Three Hundred Forty Five"
+ * 1234567 -> "One Million Two Hundred Thirty Four Thousand Five Hundred Sixty Seven"
+ * 
+ *   Did you see a pattern in dividing the number into chunk of words? For example, 123 
+ * and 123000.
+ *
+ *   Group the number by thousands (3 digits). You can write a helper function that 
+ * takes a number less than 1000 and convert just that chunk to words.
+ *
+ *   There are many edge cases. What are some good test cases? Does your code work with 
+ * input such as 0? Or 1000010? (middle chunk is zero and should not be printed out)
+ *               
+ ***************************************************************************************/
+
+#include <stdlib.h>
+#include <iostream>
+#include <vector>
+#include <string>
+using namespace std;
+
+static string dict1[] ={"Zero","One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine",
+                 "Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", 
+                 "Seventeen", "Eighteen", "Nineteen"};
+                 
+static string dict2[]={"","", "Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety" };
+static string dict3[]={"Hundred", "Thousand", "Million", "Billion" };
+
+
+// This function only converts the number which less than 1000
+string numberLess1000ToWords(int num) {
+    //char n[3] = {0, 0, 0};
+    string result;
+
+    if (num == 0) {
+        return result;
+    }else if (num < 20) {
+        return dict1[num];
+    } else if (num < 100) {
+        result = dict2[num/10];
+        if (num%10 > 0) { 
+            result += " " + dict1[num%10];
+        }
+    }else {
+        result = dict1[num/100] + " " + dict3[0];
+        if ( num % 100 > 0 ) {
+            result += " " + numberLess1000ToWords( num % 100 );
+        }
+    }
+    return result;
+}
+
+string numberToWords(int num) {
+    //edge case
+    if (num ==0 ) return dict1[num];
+
+    vector<string> ret;
+    for( ;num > 0; num/=1000 ) {
+        ret.push_back( numberLess1000ToWords(num % 1000) );
+    }
+
+    string result=ret[0];
+    for (int i=1; i<ret.size(); i++){
+        if (ret[i].size() > 0 ){
+            if ( result.size() > 0 ) {
+                result = ret[i] + " " + dict3[i] + " " + result; 
+            } else {
+                result = ret[i] + " " + dict3[i]; 
+            }
+        }
+
+    }
+    return result;
+}
+
+
+#define TEST(num) cout << num << " -> \"" << numberToWords(num) << "\"" << endl
+int main(int argc, char** argv) 
+{
+    int num = 123;
+    if (argc >1){
+        num = atoi(argv[1]);
+    }
+    TEST(num);
+
+    TEST(0);
+    TEST(1);
+    TEST(10);
+    TEST(11);
+    TEST(18);
+    TEST(20);
+    TEST(22);
+    TEST(30);
+    TEST(99);
+    TEST(100);
+    TEST(101);
+    TEST(110);
+    TEST(120);
+    TEST(256);
+    TEST(1000);
+    TEST(1001);
+    TEST(1010);
+    TEST(1110);
+    TEST(1111);
+    TEST(10000);
+    TEST(10001);
+    TEST(100000);
+    TEST(100001);
+    TEST(1000000);
+    TEST(1000001);
+    TEST(10000000);
+    TEST(10000001);
+    TEST(100000000);
+    TEST(100000001);
+    TEST(1000000000);
+    TEST(1000000001);
+    TEST(2147483647);
+
+    return 0;
+}