Python中英语单词的整数
为了解决这个问题，我们将遵循以下步骤-
定义一些列表，例如less_than_20，它将保存从1到19的所有单词
另一个数组，例如数十个，可容纳数十个，二十个，三十个，以此类推，最多可容纳九十个
数千个数组，可容纳数千，百万和十亿个
定义一个名为的函数helper()，这将花费n
如果n为0，则返回空白字符串
否则，当n<20时，返回less_than_20[n]+空白
否则，当n<100时，返回tens[n/10]+空格+helper（nmod10）
否则返回less_than_20[n/100]+“一百”+助手（nmod100）
在主要方法中，执行以下操作
如果num为0，则返回“零”
ans：=空字符串，i：=0
而num>0
ans：=helper（nummod1000）+千[i]+空格+ans
num：=num/1000
如果nummod1000不为0，则
返回ans
示例
让我们看下面的实现以更好地理解-
class Solution(object):
less_than_20 = ["", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine", "Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen",
"Nineteen"]
tens = ["","Ten", "Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety"]
thousands = ["", "Thousand", "Million", "Billion"]
def numberToWords(self, num):
if num == 0:
return "Zero"
ans = ""
i = 0
while num > 0:
if num % 1000 != 0:
ans = self.helper(num % 1000) + Solution.thousands[i] + " " + ans
i += 1
num //= 1000
return ans.strip()
def helper(self, n):
if n == 0:
return ""
elif n < 20:
return Solution.less_than_20[n] + " "
elif n < 100:
return Solution.tens[n//10] + " " + self.helper(n % 10)
else:
return Solution.less_than_20[n // 100] + " Hundred " + self.helper(n % 100)
ob = Solution()print(ob.numberToWords(512))
print(ob.numberToWords(7835271))输入值
512 7835271
输出结果
Five Hundred Twelve Seven Million Eight Hundred Thirty Five Thousand Two Hundred Seventy One