leetcode笔记：Integer to English Words

一. 题目描述

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)

二. 题目分析

题目的要求很简单，将一个整数翻译为英文说法。根据提示我们知道，数字的读法是有规律的，只需每3位作一次翻译处理即可。在循环之前，把可能用到的字符串（0-9、10-19、20、30、40、…、90等数的英语）进行归类，放在几个数组里，分情况使用。在累加字符串时注意空格的使用。另一个需要注意的是num的最大值为2,147,483,647，因此使用到的最高的单位是"Billion"。

三. 示例代码

class Solution
{
public:
    string numberToWords(int num) {
        // num的最大值：2,147,483,647
        const string unit[4] = {"", " Thousand", " Million", " Billion"};
        string result;
        int partNum[4];
        for (int i = 0; i < 4; ++i)
        {
            partNum[i] = num % 1000;
            num /= 1000;
            if (partNum[i] == 0) continue;
            result = stitch(partNum[i]) + unit[i] + result;
        }
        // 去掉result首位的空格
        return result.size() ? result.substr(1) : "Zero";
    }

private:
    string stitch(int num) {
        const string lessTen[10] = 
        {"", " One", " Two", " Three", " Four", " Five", " Six", " Seven", " Eight", " Nine"};
        const string lessTwenty[10] = 
        {" Ten", " Eleven", " Twelve", " Thirteen", " Fourteen", " Fifteen", " Sixteen",
        " Seventeen", " Eighteen", " Nineteen"};
        const string lessHundred[10] = 
        {"", "", " Twenty", " Thirty", " Forty", " Fifty", " Sixty", " Seventy", 
        " Eighty", " Ninety"};

        string temp;

        if (num != 0)
        {
            int units, tens, hundreds; // 个位、十位、百位出现的
            hundreds = num / 100;
            tens = (num % 100) / 10;
            units = (num % 100) % 10;

            if (hundreds != 0)
                temp = temp + lessTen[hundreds] + " Hundred";

            if (tens != 0)
            {
                if (tens == 1)
                {
                    temp += lessTwenty[units];
                    return temp;
                }
                else
                    temp += lessHundred[tens];
            }

            if (units != 0)
                temp += lessTen[units];
        }
        return temp;
    }
};

四. 小结

该题还是需要考虑很多边界条件的，比如数字中出现0要如何表示。