单词表示数字-LintCode

最新推荐文章于 2022-03-31 16:18:46 发布

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最新推荐文章于 2022-03-31 16:18:46 发布

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分类专栏： LintCode

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该博客介绍如何将非负整数以英文单词形式输出，例如将数字125转化为'one hundred twenty five'。文章重点在于根据数字的位数判断对应的英文表示，并采用递归方法解决位数较长的情况。

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给一个非负整数 n, 用单词打印数字

注意事项
n <= 2147483647

样例：
给出 n = 125
返回 one hundred twenty five

思路：
根据n的位数来判断是”billion”,”million”,”thousand”,”hundred”,还是十位数或者个位数。
对于位数较长的数字，递归求解。

#ifndef C688_H
#define C688_H
#include<iostream>
#include<string>
#include<vector>
using namespace std;
class Solution {
public:
    /*
    * @param : the number
    * @return: the number in words
    */
    string convertWords(int number) {
        // Write your code here
        string str = to_string(number);
        int len = str.size();
        //构建字典
        vector<string> dic{ "zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "ten",
         "eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen" };  
        vector<string> num{"","ten","twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty", "ninety"};
        string res;
        if (len == 1)
            return dic[number];
        else if (len == 2)
        {
            res += num[str[0] - '0'];
            if (res == "ten")   //特殊处理10
            {
                if (str[1] == '0')
                    return res;
                else
                    return dic[number];
            }
            else                //是否是整除的十位数
            {
                if (str[1] != '0')
                    res += " " + convertWords(number%10);
                return res;
            }
        }
        else if (len == 3)      
        {
            res += dic[str[0] - '0'] + " hundred";
            if (number % 100 != 0)
                res += " " + convertWords(number % 100);
            return res;
        }
        //4-6位后面都是thousand,递归求出?thousand,再加上百位以前的数字
        else if (len >= 4 && len <= 6)
        {
            res += convertWords(number / 1000) + " thousand";
            if (number % 1000 != 0)
                res += " " + convertWords(number % 1000);
            return res;
        }
        else if (len >= 7 && len <= 9)
        {
            res += convertWords(number / 1000000) + " million";
            if (number % 1000000 != 0)
                res += " " + convertWords(number % 1000000);
            return res;
        }
        else
        {
            res += convertWords(number / 1000000000) + " billion";
            if (number % 1000000000 != 0)
                res += " " + convertWords(number % 1000000000);
            return res;
        }
    }
};
#endif