LeetCode 273. 整数转换英文表示

1. 题目

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"

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/integer-to-english-words
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2. 解题

题目不难，对各种情况的考虑需要全面
写的不是很简洁，大家有更简洁的请告之学习观摩

class Solution {unordered_map<int,string> bits = {{1,"One"},{2,"Two"},{3,"Three"},{4,"Four"},{5,"Five"},{6,"Six"},{7,"Seven"},{8,"Eight"},{9,"Nine"},{10,"Ten"},{11,"Eleven"},{12,"Twelve"},{13,"Thirteen"},{14,"Fourteen"},{15,"Fifteen"},{16,"Sixteen"},{17,"Seventeen"},{18,"Eighteen"},{19,"Nineteen"},{20,"Twenty"},{30,"Thirty"},{40,"Forty"},{50,"Fifty"},{60,"Sixty"},{70,"Seventy"},{80,"Eighty"},{90,"Ninety"}};
public:string numberToWords(int num) {if(num == 0)return "Zero";string ans, temp;int i = 0;vector<string> units = {"","Thousand","Million","Billion"};while(num){temp = tran(num%1000);if(temp != "")//三位数都是0，跳过{temp = temp + (i>0 ? (" " + units[i]) : "");ans = " " + temp + ans;}i++;num /= 1000;//每3位一处理}if(ans.front() == ' ')    ans = ans.substr(1);//最前面可能有空格，删除return ans;}string tran(int num)//处理3位数{if(num == 0)return "";string str;int k = num/100;//百位数if(k>0)str = bits[k] + " " + "Hundred";num %= 100;//后两位if(num>0 && num <20)if(str != "")//百位不为空str += " " + bits[num];elsestr = bits[num];else if(num >= 20){k = num/10*10;// k = ？0if(k != 0){if(str != "")//前面不为0str += " " + bits[k];elsestr += bits[k];}num %= 10;//个位if(num > 0)if(str != "")//前面不为0str += " " + bits[num];elsestr += bits[num];}return str;}
};