<p>你的核心逻辑很好。基本上,您的方法将数字分解为10的幂。(数量*价值)</p>
<p>例如:2004=2*1000+0*100+0*10+4*1</p>
<p>然后转换为数量和值连接的字符串:</p>
<p>例如。
2*1000+1*4=(‘二’+‘千’)+(‘一’+‘四’)。你知道吗</p>
<p>本质上2*1000是,2<code>Thousands</code>。问题是,在最后一位小数点处,代码将1*4转换为1*4,因为从技术上讲,1*4实际上是4个1。你知道吗</p>
<p>所以检查是否在最后一个小数位,并省略值的字符串,这是我在下面的代码中所做的。你知道吗</p>
<p>有其他的边缘情况固定。例如20014是两万和十四,而不是两万,一个十和四。这是通过检查num是否在每个循环迭代的<code>store</code>中而不是只在开始时。你知道吗</p>
<pre><code>def numberToWords(num):
store = {1: 'One', 2: 'Two', 3: 'Three', 4: 'Four', 5: 'Five', 6: 'Six', 7: 'Seven', 8: 'Eight', 9: 'Nine', 11: 'Eleven', 12: 'Twelve', 13: 'Thirteen', 14: 'Fourteen', 15 : 'fifteen', 16: 'Sixteen', 17: 'Seventeen', 18: 'Eighteen',
19: 'Nineteen'}
word = ['Billion', 'Million', 'Thousand', 'Hundred', 'Ninety', 'Eighty', 'Seventy', 'Sixty', 'Fifty',\
'Forty', 'Thirty', 'Twenty', 'Ten', 'Nine', 'Eight', 'Seven', 'Six', 'Five', 'Four', 'Three', 'Two', 'One']
value = [1000000000, 1000000, 1000, 100, 90, 80, 70, 60, 50, 40, 30, 20, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1]
if num == 0: return 'Zero'
result = ''
if num in store:
return store[num]
for i in range(len(value)):
if (num //value[i] >= 1):
y = num // value[i]
if num in store: # < Check if num now has a standard word
result += store[num]
break
elif num > 9 and y > 1: # < check for last decimal place
result += numberToWords(y) + " " + word[i] + " "
elif num > 9 and y == 1:
result += word[i] + " "
else:
result += 'and ' + numberToWords(num)
num %= value[i]
return result
print(numberToWords(20014)) # Twenty Thousand Fourteen
print(numberToWords(121)) # Hundred Twenty One
</code></pre>