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

Example 1:

Input: 123

Output: "One Hundred Twenty Three"

Example 2:

Input: 12345

Output: "Twelve Thousand Three Hundred Forty Five"

Example 3:

Input: 1234567

Output: "One Million Two Hundred Thirty Four Thousand Five Hundred Sixty Seven"

Example 4:

Input: 1234567891

Output: "One Billion Two Hundred Thirty Four Million Five Hundred Sixty Seven Thousand Eight Hundred Ninety

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Hint:

Did you see a pattern in dividing the number into chunk of words? For example, 123 and 123000.
Group the number by thousands (3 digits). You can write a helper function that takes a number less than 1000 and convert just that chunk to words.
There are many edge cases. What are some good test cases? Does your code work with input such as 0? Or 1000010? (middle chunk is zero and should not be printed out)

将一个整数转换成英文单词。提示：要3个一组进行处理，限定了数字范围为0到2^31 - 1之间，最高只能到billion位，只需处理四组。

每三位一组进行处理，每加一组就加上units，整数除以1000，对1000取余。处理范围分为99以上，20~99，10~20，0~10。先写出不同的数字范围英文单词列表，然后根据范围添加到结果中。

F家：可能是负数

Java:

class Solution {

    public String numberToWords(int num) {

            String[] units = {""," Thousand"," Million"," Billion"};

            int i = 0;

            String res="";

            while(num > 0) {

                int temp = num % 1000;

                if(temp > 0) res = convert(temp) + units[i] + (res.length()==0 ?"": " "+res);

                num /= 1000;

                i++;

            }

            return res.isEmpty()? "Zero" : res;

        }

        public String convert(int num){

            String res = "";

            String[] ten = {"Zero", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine"};

            String[] hundred = {"Ten", "Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety"};

            String[] twenty = {"Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen"};

            if(num>0) {

                int temp = num / 100;

                if(temp > 0) {

                    res += ten[temp] + " Hundred";

                }

                temp = num%100;

                if(temp >= 10 && temp < 20){

                    if(!res.isEmpty()) res +=" ";

                    res = res + twenty[temp%10];

                    return res;

                }else if(temp >= 20){

                    temp = temp / 10;

                    if(!res.isEmpty()) res +=" ";

                    res = res + hundred[temp-1];

                }

                temp = num % 10;

                if(temp > 0) {

                    if(!res.isEmpty()) res +=" ";

                    res = res + ten[temp];

                }

            }

            return res;

        }

    }

}

Python:

class Solution(object):

    def numberToWords(self, num):

        """

        :type num: int

        :rtype: str

        """

        if num == 0:

            return "Zero"

        lookup = {0: "Zero", 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", \

                  20: "Twenty", 30: "Thirty", 40: "Forty", 50: "Fifty", 60: "Sixty", \

                  70: "Seventy", 80: "Eighty", 90: "Ninety"}

        unit = ["", "Thousand", "Million", "Billion"]

        res, i = [], 0

        while num:

            cur = num % 1000

            if num % 1000:

                res.append(self.threeDigits(cur, lookup, unit[i]))

            num //= 1000

            i += 1

        return " ".join(res[::-1])

    def threeDigits(self, num, lookup, unit):

        res = []

        if num / 100:

            res = [lookup[num / 100] + " " + "Hundred"]

        if num % 100:

            res.append(self.twoDigits(num % 100, lookup))

        if unit != "":

            res.append(unit)

        return " ".join(res)

    def twoDigits(self, num, lookup):

        if num in lookup:

            return lookup[num]

        return lookup[(num / 10) * 10] + " " + lookup[num % 10]

Python:

class Solution(object):

    def numberToWords(self, num):

        lv1 = "Zero One Two Three Four Five Six Seven Eight Nine Ten \

               Eleven Twelve Thirteen Fourteen Fifteen Sixteen Seventeen Eighteen Nineteen".split()

        lv2 = "Twenty Thirty Forty Fifty Sixty Seventy Eighty Ninety".split()

        lv3 = "Hundred"

        lv4 = "Thousand Million Billion".split()

        words, digits = [], 0

        while num:

            token, num = num % 1000, num / 1000

            word = ''

            if token > 99:

                word += lv1[token / 100] + ' ' + lv3 + ' '

                token %= 100

            if token > 19:

                word += lv2[token / 10 - 2] + ' '

                token %= 10

            if token > 0:

                word += lv1[token] + ' '

            word = word.strip()

            if word:

                word += ' ' + lv4[digits - 1] if digits else ''

                words += word,

            digits += 1

        return ' '.join(words[::-1]) or 'Zero'

C++:

class Solution {

public:

    string numberToWords(int num) {

        string res = convertHundred(num % 1000);

        vector<string> v = {"Thousand", "Million", "Billion"};

        for (int i = 0; i < 3; ++i) {

            num /= 1000;

            res = num % 1000 ? convertHundred(num % 1000) + " " + v[i] + " " + res : res;

        }

        while (res.back() == ' ') res.pop_back();

        return res.empty() ? "Zero" : res;

    }

    string convertHundred(int num) {

        vector<string> v1 = {"", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine", "Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen"};

        vector<string> v2 = {"", "", "Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety"};

        string res;

        int a = num / 100, b = num % 100, c = num % 10;

        res = b < 20 ? v1[b] : v2[b / 10] + (c ? " " + v1[c] : "");

        if (a > 0) res = v1[a] + " Hundred" + (b ? " " + res : "");

        return res;

    }

};

All LeetCode Questions List 题目汇总

巴特西

[LeetCode] 273. Integer to English Words 整数转为英文单词

All LeetCode Questions List 题目汇总

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