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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"
Hint:
一定要考虑各种边界情况。正如提示里所说的。
1 class Solution { 2 public: 3 string getWord(int num, string *digit, string *digit1) { 4 if (num == 0) return "Zero"; 5 string res; 6 int h = num / 100; 7 num %= 100; 8 if (h > 0) res += digit[h] + " Hundred"; 9 if (num == 0) return res; 10 else if (h > 0) res += " "; 11 if (num < 20) { 12 res += digit[num]; 13 } else { 14 h = num / 10; 15 num %= 10; 16 res += digit1[h]; 17 if (num != 0) res += " " + digit[num]; 18 } 19 return res; 20 } 21 string numberToWords(int num) { 22 if (num == 0) return "Zero"; 23 string digit[20] = {"Zero", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine", 24 "Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen" 25 }; 26 string digit1[10] = {"", "", "Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety"}; 27 string digit2[10] = {"", "Thousand", "Million", "Billion"}; 28 vector<string> words; 29 string str; 30 int num2, cnt = 0; 31 while (num != 0) { 32 num2 = num % 1000; 33 str = getWord(num2, digit, digit1); 34 if (str != "Zero") { 35 if (cnt > 0) words.push_back(str + " " + digit2[cnt]); 36 else words.push_back(str); 37 } 38 num /= 1000; 39 ++cnt; 40 } 41 string res; 42 for (int i = (int)words.size() - 1; i > 0; --i) { 43 res += words[i] + " "; 44 } 45 res += words.front(); 46 return res; 47 } 48 };
[LeetCode] Integer to English Words
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原文地址：http://www.cnblogs.com/easonliu/p/4772781.html
