【LeetCode】273. 整数转换英文表示 结题报告 (C++)

原题地址：https://leetcode-cn.com/problems/integer-to-english-words/

题目描述：

将非负整数转换为其对应的英文表示。可以保证给定输入小于 231 - 1 。

示例 1:

输入: 123
输出: "One Hundred Twenty Three"
示例 2:

输入: 12345
输出: "Twelve Thousand Three Hundred Forty Five"
示例 3:

输入: 1234567
输出: "One Million Two Hundred Thirty Four Thousand Five Hundred Sixty Seven"
示例 4:

输入: 1234567891
输出: "One Billion Two Hundred Thirty Four Million Five Hundred Sixty Seven Thousand Eight Hundred Ninety One"

解题方案：

我又懒惰了，看着这题麻烦又不想做了。

扫描二维码关注公众号，回复： 4367275 查看本文章

参考地址：https://blog.csdn.net/Bendaai/article/details/82284104?utm_source=blogxgwz8

本题更像是一道设计题目，首先先定义好各种数字的英文单词，之后就是对数字进行拆分，一直做除法。注意几个关键数字：0，20，100 。

class Solution {
public:
    const int Mod[3] = { 1000000000,1000000,1000 };
    string H[3] = { "Billion","Million","Thousand" },
        M[8] = { "Twenty","Thirty","Forty","Fifty","Sixty","Seventy","Eighty","Ninety" },
        L[20] = { "Zero","One","Two","Three","Four","Five","Six","Seven","Eight","Nine","Ten",
        "Eleven","Twelve","Thirteen","Fourteen","Fifteen","Sixteen","Seventeen","Eighteen","Nineteen" };
    void update(string &ans) {
        ans += ans == "" ? "" : " ";
    }
    string numberToWords2(int num) {//num小于1000
        if (num < 20) 
            return L[num];//zero的情况
        string ans;
        if (num >= 100) 
            ans += L[num / 100] + " Hundred",num%=100;//num变两位数
        if (num == 0) 
            return ans;
        else if (num < 20) 
            update(ans),ans += L[num];
        else {
            update(ans), ans += M[num / 10 - 2], num %= 10;//num变一位数
            if (num == 0) 
                return ans;
            else  
                update(ans), ans+= L[num];
        }
        return ans;
    }
    string numberToWords(int num) {
        if (num < 20) //zero的情况
            return L[num];
        string ans;
        for(int i = 0; i < 3; ++i)
            if (num >= Mod[i]) 
                update(ans),ans += numberToWords2(num / Mod[i]) + " " + H[i], num %= Mod[i];
        if(num) 
            update(ans), ans += numberToWords2(num);
        return ans;
    }
};