Question

首先，我是python和一般编程的初学者。我正在尝试解决将正整数转换为单词的leetcode问题。我的算法似乎几乎是正确的，但存在一些瓶颈。我不知道如何摆脱数量少于一百的one，请查看输出，这将导致转换错误。我已经有一段时间没有运气了，以为我应该寻求帮助。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

您的核心逻辑很好。本质上，您的方法将数字分解为十的幂。（数量*值）

例如2004 = 2 * 1000 + 0 * 100 + 0 * 10 + 4 * 1

然后将其转换为具有连接的数量和值的字符串：

例如 2 * 1000 + 1 * 4 =（'两个'+'千'）+（'一个'+'四'）。

基本上2 * 1000是两个Thousands。错误是，在最后一个小数点后，您的代码将1 * 4转换为四，因为从技术上讲1 * 4实际上是四个。

因此，请检查您是否位于小数点后一位，并省略该值的字符串，这在下面的代码中已经完成。

还有其他一些固定的情况。例如，20014是2万和14，而不是2万，10和4。这是通过检查num是否在每次循环迭代中都位于store中而不是仅在开始时进行的。

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

Answer 2

如果value [i]小于100，则不要打印一个。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))

输出：123.345万

python中整数到单词的错误答案

2 个答案: