[LeetCode 273] Integer to English Word

[LeetCode 273] Integer to English Word

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

Hint:

1. Did you see a pattern in dividing the number into chunk of words? For example, 123 and 123000.
2. 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.
3. 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) solution:

   Just as hint described, handle 3digit at each time, convert them into English word,

   every time handle 3 digit, the unit will increase.

   So many corner cases need to handle.

    

   public String numberToWords(int num) {
           String[] units = {, Thousand, Million, Billion};
           int i = 0;
           String res=;
           while(num>0) {
               int temp = num%1000;
               if(temp>0) res = convert(temp) + units[i] + (res.length()==0 ?:  +res);
               num /= 1000;
               i++;
           }
           return res.isEmpty()? Zero : res;
       }
       public String convert(int num){
           String res = ;
           String[] ten = {Zero, One, Two, Three, Four, Five, Six, Seven, Eight, Nine};
           String[] hundred = {Ten, Twenty, Thirty, Forty, Fifty, Sixty, Seventy, Eighty, Ninety};
           String[] twenty = {Ten, Eleven, Twelve, Thirteen, Fourteen, Fifteen, Sixteen, Seventeen, Eighteen, Nineteen};
           if(num>0) {
               int temp = num/100;
               if(temp>0) {
                   res += ten[temp] +  Hundred;
               }
               temp = num%100;
               if(temp>=10 && temp<20){
                   if(!res.isEmpty()) res += ;
                   res = res + twenty[temp%10];
                   return res;
               }else if(temp>=20){
                   temp = temp/10;
                   if(!res.isEmpty()) res += ;
                   res = res + hundred[temp-1];
               }
               temp = num%10;
               if(temp>0) {
                   if(!res.isEmpty()) res += ;
                   res = res + ten[temp];
               }
           }
           return res;
       }