273. Integer to English Words

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题目：

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)

链接： http://leetcode.com/problems/integer-to-english-words/

题解：

把数字翻译成英文。这个跟Integer to Roman很像，把情况分清楚就不难解决。代码大都参考了Discuss里hwy_2015的，简洁易懂。主要是以1000为一个单位来把数组分成组，每个组内单独处理tens和lessThanTwenty的情况。

Time Complexity - O(n)， Space Complexity - O(n)。

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

Reference:

https://leetcode.com/discuss/55462/my-clean-java-solution-very-easy-to-understand

https://leetcode.com/discuss/55349/if-you-know-how-to-read-numbers-you-can-make-it

https://leetcode.com/discuss/71544/short-clean-java-solution

https://leetcode.com/discuss/60010/share-my-clean-java-solution

https://leetcode.com/discuss/55273/my-java-solution

https://leetcode.com/discuss/55477/recursive-python

273. Integer to English Words

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原文地址：http://www.cnblogs.com/yrbbest/p/5031682.html