273. Integer to English Words

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本文介绍了一种将非负整数转换为其英文单词表示的方法，适用于小于2^31 - 1的整数。通过使用预定义的数组来处理个位、十位、百位及千位以上的数字，并通过递归地分解数字来实现转换。

问题链接：https://leetcode.com/problems/integer-to-english-words/?tab=Description

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:

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)

想了半天还是没什么思路，参照答案默写如下：

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