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。
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;
}
};