python中整数到单词的错误答案

Question

首先，我是python和编程的初学者。我想解决一个leetcode问题，把正整数转换成word。我的算法似乎几乎是正确的，但有一些瓶颈。 我不知道如何去掉小于100的one，请看输出，这使得转换错误。 我干了一段时间，运气不好，觉得应该寻求帮助。谢谢

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]
            result += numberToWords(y) + " " + word[i] + " "
        num %= value[i]

    return result

print(numberToWords(12345))

我的输出："Twelve Thousand Three Hundred One Forty One Five " 预期输出："Twelve Thousand Three Hundred Forty Five"

Answer 1

如果值[i]小于100，就不要打印一个。在这里：

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 value[i] < 100:
               result += word[i] + " "
            else:
               result+=numberToWords(y) + " " + word[i] + " "
        num %= value[i]

    return result

print(numberToWords(12345))

产量：一万二千三百四十五

Answer 2

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）

例如：2004=2*1000+0*100+0*10+4*1

然后转换为数量和值连接的字符串：

例如。 2*1000+1*4=（‘二’+‘千’）+（‘一’+‘四’）。你知道吗

本质上2*1000是，2Thousands。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

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

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store中而不是只在开始时。你知道吗

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