LeetCode #273: Integer to English Words

Problem Statement

(Source) Convert a non-negative integer to its english words representation. Given input is guaranteed to be less than 231−1.

For example,

 
123 -> “One Hundred Twenty Three” 
12345 -> “Twelve Thousand Three Hundred Forty Five” 
1234567 -> “One Million Two Hundred Thirty Four Thousand Five Hundred Sixty Seven” 

Analysis

Split the input number into groups of 3 digits from lowest bit to highest bit. Then process each group and add corresponding units to the result of each group. Combine sub-answer to get the final global answer.

Solution

class Solution(object):    def process(self, map1, map2, num):        """Process a non-negtive number which contains no more than 3 digits.        """        if num == 0:            return ''        elif num < 20:            return map1[num]        elif num >= 20 and num <= 99:            ans = map2[num / 10]            if num % 10:                ans = ans + ' ' + map1[num % 10]            return ans        else:            ans = map1[num / 100] + ' Hundred'            if num % 100:                ans = ans + ' ' + self.process(map1, map2, num % 100)            return ans    def numberToWords(self, num):        """        :type num: int        :rtype: str        """        if num == 0:            return 'Zero'        res = []        map1 = {1: 'One', 2: 'Two', 3: 'Three', 4: 'Four', 5: 'Five', 6: 'Six', 7: 'Seven', 8: 'Eight', 9: 'Nine', 10: 'Ten', 11: 'Eleven', 12: 'Twelve', 13: 'Thirteen', 14: 'Fourteen', 15: 'Fifteen', 16: 'Sixteen', 17: 'Seventeen', 18: 'Eighteen', 19: 'Nineteen'}        map2 = {2: 'Twenty', 3: 'Thirty', 4: 'Forty', 5: 'Fifty', 6: 'Sixty', 7: 'Seventy', 8: 'Eighty', 9: 'Ninety'}        unit_map = {1: 'Thousand', 2: 'Million', 3: 'Billion'}        unit = 0        while num:            x = num % 1000            sub_ans = self.process(map1, map2, x)            if sub_ans:                if unit:                    sub_ans = sub_ans + ' ' + unit_map[unit]                res.append(sub_ans)            num /= 1000            unit += 1        res.reverse()        return ' '.join(res)