假设我们有一个数字。数字可以是0到231 – 1之间的任何数字。我们必须将数字转换为单词。因此，如果数字是512，那么结果将是512。
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
数千个数组，可容纳数千，百万和十亿个
定义一个名为helper()的函数，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（n mod 10）
否则返回less_than_20 [n / 100] +“一百” +助手（n mod 100）
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（num mod 1000）+千[i] +空格+ ans
如果num mod 1000不为0，则
返回ans
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))512
7835271输出结果
Five Hundred Twelve
Seven Million Eight Hundred Thirty Five Thousand Two Hundred Seventy One