LeetCode 273. Integer to English Words

I changed the integer into string. If the length is not multiple of 3, append "0" ahead. Then, deal three digits by chunks.

#include <iostream>
#include <string>
#include <vector>
using namespace std;

/*
  Convert a non-negative integer to its English words representation.
  Given input is guaranteed to be less then 2 ^ 31 - 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"

  Seems the best way is to cut the number by 3-digits.
  1234567 --> 1, 234, 567
*/

vector<string> digit_1{"", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine", "Ten", "Eleven", "Tweleve", "Thirteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen"};
vector<string> digit_2{"Twenty", "Thrity", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety", "Hundred"};
vector<string> digit_3{"", "Thousand", "Million", "Billion"};

string threeDigits(string tmp) {
  string res = "";
  int n = tmp.size();
  if(n == 1) return digit_1[tmp[2] - '0'];
  else if(n == 2){
    if(tmp[1] >= '2') return res = digit_2[tmp[1] - '0' - 2] + " " + digit_1[tmp[2] - '0'];
    else {
      int number = (tmp[1] - '0') * 10 + tmp[2] - '0';
      return res = digit_1[number];
    }
  } else {
    if(tmp[0] != '0') {
      res = res + digit_1[tmp[0] - '0'] + " " + "Hundred";
      if(tmp[1] >= '2') res = res + " " + digit_2[tmp[1] - '0' - 2] + " " + digit_1[tmp[2] - '0'];
      else {
        int number = (tmp[1] - '0') * 10 + tmp[2] - '0';
        res += " " + digit_1[number];
      }
    }
    else {
      if(tmp[1] >= '2') res = res + digit_2[tmp[1] - '0' - 2] + " " + digit_1[tmp[2] - '0'];
      else {
        int number = (tmp[1] - '0') * 10 + tmp[2] - '0';
        res += " " + digit_1[number];
      }
    }
  }
  return res;
}
string numberToWords(int num) {
  string tmp = to_string(num);
  int i = 0;
  int count = 0;
  if(tmp.size() % 3) count = 3 - (tmp.size() % 3);
  string res = "";
  for(int k = 0; k < count; ++k) {
    tmp = "0" + tmp;
  }
  int digits = tmp.size() / 3;
  count = tmp.size() / 3;
  while(i + 3 <= tmp.size()) {
    string a  = threeDigits(tmp.substr(i, 3)) + " " + digit_3[--count];
    i = i + 3;
    res = res + " " + a;
  }
  return res;
}

int main(void) {
  string res = numberToWords(1200);
  cout << res << endl;
}

The above way is easy to come up but hard to be bug free.

The following code is more cleaner.

class Solution {
public:
    static string numberToWords(int n) {
        if(n == 0) return "Zero";
        else return int_string(n).substr(1);
    }
private:
    static const char * const below_20[];
    static const char * const below_100[];
    static string int_string(int n) {
        if(n >= 1000000000)   return int_string(n / 1000000000) + " Billion" + int_string(n - 1000000000 * (n / 1000000000));
        else if(n >= 1000000) return int_string(n / 1000000) + " Million" + int_string(n - 1000000 * (n / 1000000));
        else if(n >= 1000)    return int_string(n / 1000) + " Thousand" + int_string(n - 1000 * (n / 1000));
        else if(n >= 100)     return int_string(n / 100) + " Hundred" + int_string(n - 100 * (n / 100));
        else if(n >= 20)      return string(" ") + below_100[n / 10 - 2] + int_string(n - 10 * (n / 10));
        else if(n >= 1)       return string(" ") + below_20[n - 1];
        else return "";
        }
    }
};

const char * const Solution::below_20[] =  {"One", "Two", "Three", "Four","Five","Six","Seven","Eight","Nine","Ten", "Eleven","Twelve","Thirteen","Fourteen","Fifteen","Sixteen","Seventeen","Eighteen","Nineteen"};
const char * const Solution::below_100[] = {"Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety"};