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"

思路

1. 题目首先告诉我们要3个一组的进行处理，而且限定了输入数字范围为0到231 - 1之间，最高只能到billion位，3个一组也只需处理四组即可，那么我们需要一个处理三个一组数字的函数。
2. 需要把1到9，10 - 19的英文单词都列出来，分别放到2个数组里；还要把20,30，... 到90的英文单词列出来放到另一个数组里。
3. 然后我们需要用技巧，比如一个三位数n
   • 百位数表示为n/100
   • 后两位数一起表示为n%100
   • 十位数表示为n%100/10
   • 个位数表示为n%10
4. 然后我们看后两位数是否小于10 或者小于20，小于的话直接从数组中取出单词
5. 如果大于等于20的话，则分别将十位和个位数字的单词从两个数组中取出来。
6. 然后再来处理百位上的数字，还要记得加上Hundred。然后中间要插入"Thousand", "Million", "Billion"到对应的位置
7. 最后check一下末尾是否有空格，把空格都删掉

Coner Case: 检查下输入是否为0 或者小于0，是的话要返回'Zero' 或者NULL。

class Solution {
    String[] belowTen = {"","One","Two","Three","Four","Five","Six","Seven","Eight","Nine"};
        String[] teens = {"Ten", "Eleven", "Twelve","Thirteen","Fourteen","Fifteen","Sixteen","Seventeen","Eighteen","Nineteen"};
        String[] belowHundreds = {"","Ten","Twenty","Thirty","Forty","Fifty","Sixty","Seventy","Eighty","Ninety"};
        
    public String numberToWords(int num) {
        if (num < 0) return null;
        if (num == 0) return "Zero";
        return helper(num);
    }
    
    public String helper(int num) {
        String result = new String();
        if (num < 10) {
            result = belowTen[num];
        } else if (num < 20) {
            result = teens[num % 10];
        } else if (num < 100) {
            result = belowHundreds[num / 10] + " " + belowTen[num % 10];
        } else if (num < 1000) {
            result = belowTen[num / 100] + " Hundred " + helper(num % 100);
        } else if (num < 1000000) {
            result = helper(num / 1000) + " Thousand " + helper(num % 1000);
        } else if (num < 1000000000) {
            result = helper(num / 1000000) + " Million " + helper(num % 1000000);
        } else {
            result = helper(num / 1000000000) + " Billion " + helper(num % 1000000000);
        }
        return result.trim();
    }
}