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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:
这道题让我们把一个整型数转为用英文单词描述,就像在check上写钱数的方法,我最开始的方法特别复杂,因为我用了几个switch语句来列出所有的单词,但是我看网上大神们的解法都是用数组来枚举的,特别的巧妙而且省地方,膜拜学习中。题目中给足了提示,首先告诉我们要3个一组的进行处理,而且题目中限定了输入数字范围为0到231 - 1之间,最高只能到billion位,3个一组也只需处理四组即可,那么我们需要些一个处理三个一组数字的函数,我们需要把1到19的英文单词都列出来,放到一个数组里,还要把20,30,... 到90的英文单词列出来放到另一个数组里,然后我们需要用写技巧,比如一个三位数n,百位数表示为n/100,后两位数一起表示为n%100,十位数表示为n%100/10,个位数表示为n%10,然后我们看后两位数是否小于20,小于的话直接从数组中取出单词,如果大于等于20的话,则分别将十位和个位数字的单词从两个数组中取出来。然后再来处理百位上的数字,还要记得加上Hundred。主函数中调用四次这个帮助函数,然后中间要插入"Thousand", "Million", "Billion"到对应的位置,最后check一下末尾是否有空格,把空格都删掉,返回的时候检查下输入是否为0,是的话要返回‘Zero‘。参见代码如下:
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; } };
参考资料:
https://leetcode.com/discuss/55268/short-clean-c-code-with-explanation
LeetCode All in One 题目讲解汇总(持续更新中...)
[LeetCode] Integer to English Words 整数转为英文单词
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原文地址:http://www.cnblogs.com/grandyang/p/4772780.html