2019-02-05
273. Integer to English Words.jpg
LeetCode 273. Integer to English Words
Description
Convert a non-negative integer to its english words representation. Given input is guaranteed to be less than 231 - 1.
Example 1:
Input: 123
Output: "One Hundred Twenty Three"
Example 2:
Input: 12345
Output: "Twelve Thousand Three Hundred Forty Five"
Example 3:
Input: 1234567
Output: "One Million Two Hundred Thirty Four Thousand Five Hundred Sixty Seven"
Example 4:
Input: 1234567891
Output: "One Billion Two Hundred Thirty Four Million Five Hundred Sixty Seven Thousand Eight Hundred Ninety One"
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"
思路
- 数字三个为一节，我们每三个数字做一个分段，对其用用英文表达出来，在加上对应节所在的单位即可.
# -*- coding: utf-8 -*-
# @Author: 何睿
# @Create Date: 2019-02-04 19:25:43
# @Last Modified by: 何睿
# @Last Modified time: 2019-02-04 20:31:32
class Solution:
def numberToWords(self, num):
"""
:type num: int
:rtype: str
"""
# 维护一个数字到英语的转换字典
self.IntEng = {
"0": 'Zero',
"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',
"100": 'Hundred',
}
# 没三节为一个单位，2^31最大用到的单位为Billion
units = ['', 'Thousand', 'Million', 'Billion']
res = ''
# 将数字转换成为字符串
_str = str(num)
# count:字符串中字符个数，i:数字三个为一节，当前数字所在节
count, i = len(_str), 0
while count - (i + 1) * 3 >= 0:
# 取出三个数字
_num = _str[count - (i + 1) * 3:count - i * 3]
# 三个数字都不为0进行计算
if not _num == "000":
s = self.__readthree(_num)
res = s + " " + units[i] + " " + res
# 进入下一节
i += 1
# 如果最后一节不足三个数，处理剩下的数
if count - (i * 3):
s = self.__readthree(_str[0:count - (i * 3)])
res = s + " " + units[i] + " " + res
# 返回结果，去掉最后的空格
return res.strip()
# 私有函数，读一节（三个数字）数
def __readthree(self, _str):
_str = str(int(_str))
if len(_str) == 1:
return self.IntEng[_str]
if len(_str) == 2:
return self.__readtwo(_str)
if len(_str) == 3:
res = self.IntEng[_str[0]] + " " + "Hundred"
if int(_str) % 100:
return res + " " + self.__readtwo(_str[1:])
else:
return res
# 私有函数，读两位数，辅助读一节数字
def __readtwo(self, _str):
_str = str(int(_str))
if int(_str) <= 20 or _str[1] == "0":
return self.IntEng[_str]
else:
return self.IntEng[_str[0] + '0'] + " " + self.IntEng[_str[1]]
