LeetCode 273. Integer to English Words

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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;
}

LeetCode 273. Integer to English Words

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原文地址：http://blog.csdn.net/github_34333284/article/details/51360235