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:
解法：按billion、million、thousand和hundred及以下依次处理即可。
class Solution { public: string numberToWords(int num) { vector<string> unit = { "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine" }; vector<string> dec1 = { "Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen" }; vector<string> dec2 = { "", "", "Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety" }; vector<string> expr = { "Thousand", "Million", "Billion" }; string res = ""; for (int i = 9; i >= 0; i -= 3) { int div = num / (int)pow(10, i); if (div > 0 && div < 10) { res += unit[div - 1] + " "; if (i > 0) res += expr[i / 3 - 1] + " "; } else if (div >= 10 && div < 20) { res += dec1[div - 10] + " "; if (i > 0) res += expr[i / 3 - 1] + " "; } else if (div >= 20) { if (div >= 20 && div < 100) { int bas = div / 10, reu = div % 10; res += dec2[bas] + " "; if (reu > 0) res += unit[reu - 1] + " "; if (i > 0) res += expr[i / 3 - 1] + " "; } else if (div >= 100) { int hun = div / 100, tne = div / 10 % 10, uni = div % 10; res += unit[hun - 1] + " Hundred "; if (tne == 1) res += dec1[div % 100 - 10] + " "; else if (tne > 1 || uni > 0) { if(tne > 1) res += dec2[tne] + " "; if (uni > 0) res += unit[uni - 1] + " "; } if (i > 0) res += expr[i / 3 - 1] + " "; } } num %= (int)pow(10, i); if (num <= 0) break; } if (res.empty()) res += "Zero "; res.resize(res.size() - 1); return res; } };
改进：将输入整数用billion、million、thousand除后，所得的商均在0-999之间，因此可以将0-999之间的读法单独提出来，作为一个工具函数。
class Solution { public: string numberToWords(int num) { if (num == 0) return "Zero"; vector<string> expr = { "Thousand", "Million", "Billion" }; string res = ""; for (int i = 9; i >= 0; i -= 3) { int div = num / (int)pow(10, i); if (div > 0) { res += lessThanThousand(div); if (i > 0) res += expr[i / 3 - 1] + " "; } num %= (int)pow(10, i); } res.resize(res.size() - 1); return res; } private: string lessThanThousand(int num) { vector<string> unit = { "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine" }; vector<string> decd = { "Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen", "Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety" }; string res = ""; int hun = num / 100, ten = num / 10 % 10, uni = num % 10; if (hun > 0) res += unit[hun - 1] + " Hundred "; if (ten == 1) res += decd[ten * 10 + uni - 10] + " "; else if (ten > 1 || uni > 0) { if (ten > 1) res += decd[8 + ten] + " "; if (uni > 0) res += unit[uni - 1] + " "; } return res; } };
[LeetCode]47. Integer to English Words整数的读法
原文：http://www.cnblogs.com/aprilcheny/p/4919849.html