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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思路：

根据英文的表达习惯，数字被分为三个三个一组，一般以逗号隔开，比如1，000，000，000。

所以不难发现本题的核心目标就是，将1000以下的整数转换成英文的代码写出来，然后将数字每三个一组，不断重复调用转换代码，最后将答案拼接在一起即可。

难不是很难，但是特别繁琐，注意英语拼写，注意edge case。

class Solution(object): def numberToWords(self, num): """ :type num: int :rtype: str """ def helper(num): #本函数用于处理1000 以下的整数转英文 n = int(num) num = str(n) if n < 100: return subhelper(num) else: return ["One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine"][int(num[0]) - 1] + " Hundred " + subhelper(num[1:]) if num[1:] != "00" else ["One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine"][int(num[0]) - 1] + " Hundred" def subhelper(num): #本函数用于处理100 以下的整数转英文 n = int(num) l1 = ["Zero", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine"] l2 = ["Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen"] l3 = ["Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety"] if n < 10: return l1[int(num)] if 10 <= n < 20: return l2[n - 10] if 20 <= n < 100: return l3[int(num[0]) - 2] + " " + l1[int(num[1])] if num[1] != "0" else l3[int(num[0]) - 2] res = "" if num >= 1000000000: res = helper(str(num)[0]) + " Billion" if str(num)[1:4] != "000": res += " " + helper(str(num)[1:4]) + " Million" if str(num)[4:7] != "000": res += " " + helper(str(num)[4:7]) + " Thousand" if str(num)[7:] != "000": res += " " + helper(str(num)[7:]) elif num >= 1000000: res = helper(str(num)[:-6]) + " Million" if str(num)[-6:-3] != "000": res += " " + helper(str(num)[-6:-3]) + " Thousand" if str(num)[-3:] != "000": res += " " + helper(str(num)[-3:]) elif num >= 1000: res = helper(str(num)[:-3]) + " Thousand" if str(num)[-3:] != "000": res += " " + helper(str(num)[-3:]) else: return helper(str(num)) return res