LeetCode273整数转英文

最新推荐文章于 2021-10-12 09:24:07 发布

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最新推荐文章于 2021-10-12 09:24:07 发布

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分类专栏： LeetCode

本文链接：https://blog.youkuaiyun.com/qq_35654080/article/details/82558650

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本文介绍了一种将非负整数转换为英文表述的方法。通过递归地将数字分为千位组并转换每一组，实现了从1到2^31 - 1范围内数字的完整英文表达。特别关注了10至19的特殊读法。

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将非负整数转换为其对应的英文表示。可以保证给定输入小于 231 - 1 。

示例 1:

输入: 123
输出: "One Hundred Twenty Three"

示例 2:

输入: 12345
输出: "Twelve Thousand Three Hundred Forty Five"

示例 3:

输入: 1234567
输出: "One Million Two Hundred Thirty Four Thousand Five Hundred Sixty Seven"

示例 4:

输入: 1234567891
输出: "One Billion Two Hundred Thirty Four Million Five Hundred Sixty Seven Thousand Eight Hundred Ninety One"

按照英文习惯将数字从低位到高位每三位划分，然后从高位到低位读数，单位分别为Billion Million Thousand;
另外注意一下10-19的读法即可

class Solution:
    def numberToWords(self, num):
        """
        :type num: int
        :rtype: str
        """
        if (num == 0):
            return "Zero"
        result = "";
        if (num > 999999999):
            result += self.numberToWordsWithThreeDIgit(num // 1000000000) + " Billion"
            num %= 1000000000

        if (num > 999999):
            result += self.numberToWordsWithThreeDIgit(num // 1000000) + " Million"
            num %= 1000000

        if (num > 999):
            result += self.numberToWordsWithThreeDIgit(num // 1000) + " Thousand"
            num %= 1000

        if (num > 0):
            result += self.numberToWordsWithThreeDIgit(num)

        return result

    def numberToWordsWithThreeDIgit(self, num):
        num1 = ["One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine"]
        num2 = ["Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen",
                "Nineteen"]
        tens = ["Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety"]
        result = ""
        if (num > 99):
            result += " " + num1[num // 100 - 1] + " Hundred"
            num %= 100
        if (num > 19):
            result += " " + tens[num // 10 - 2]
            num %= 10
        if (num > 9):
            result += " " + num2[num - 10]
            num = 0

        if (num > 0):
            result += " " + num1[num - 1]

        return result

try1=Solution()
print(try1.numberToWords(5456467))