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题目：（困难）

标签：字符串、分治算法

LeetCode的Python执行用时随缘，只要时间复杂度没有明显差异，执行用时一般都在同一个量级，仅作参考意义。

解法一：

class Solution:    def numberToWords(self, num: int) -> str:        # 定义字典        D1 = {
   1: "One", 2: "Two", 3: "Three", 4: "Four", 5: "Five", 6: "Six", 7: "Seven", 8: "Eight", 9: "Nine"}        D2 = {
   10: "Ten", 11: "Eleven", 12: "Twelve", 13: "Thirteen", 14: "Fourteen", 15: "Fifteen", 16: "Sixteen", 17: "Seventeen", 18: "Eighteen",              19: "Nineteen"}        D3 = {
   2: "Twenty", 3: "Thirty", 4: "Forty", 5: "Fifty", 6: "Sixty", 7: "Seventy", 8: "Eighty", 9: "Ninety"}        # 处理两位数以内的情况        def get_0_99(n):            if n == 0:                return []            elif n < 10:                return [D1[n]]            elif n < 20:                return [D2[n]]            else:                a, b = divmod(n, 10)                if b == 0:                    return [D3[a]]                else:                    return [D3[a], D1[b]]        # 处理三位数以内的情况        def get_0_999(n):            a, b = divmod(n, 100)            if n < 100:                return get_0_99(b)            else:                return [D1[a], "Hundred"] + get_0_99(b)        # 处理万亿以内的情况        ans = []        if num >= 1000000000:            now, num = divmod(num, 1000000000)            ans += get_0_999(now) + ["Billion"]        if num >= 1000000:            now, num = divmod(num, 1000000)            ans += get_0_999(now) + ["Million"]        if num >= 1000:            now, num = divmod(num, 1000)            ans += get_0_999(now) + ["Thousand"]        ans += get_0_999(num)        return " ".join(ans) if ans else "Zero"if __name__ == "__main__":    print(Solution().numberToWords(0))  # "Zero"    print(Solution().numberToWords(20))  # "Twenty"    print(Solution().numberToWords(123))  # "One Hundred Twenty Three"    print(Solution().numberToWords(12345))  # "Twelve Thousand Three Hundred Forty Five"    print(Solution().numberToWords(1234567))  # "One Million Two Hundred Thirty Four Thousand Five Hundred Sixty Seven"    print(Solution().numberToWords(        1234567891))  # "One Billion Two Hundred Thirty Four Million Five Hundred Sixty Seven Thousand Eight Hundred Ninety One"

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