[LeetCode] 273. Integer to English Words 整数转为英文单词

Convert a non-negative integer num to its English words representation.

Example 1:

Input: num = 123
Output: "One Hundred Twenty Three"

Example 2:

Input: num = 12345
Output: "Twelve Thousand Three Hundred Forty Five"

Example 3:

Input: num = 1234567
Output: "One Million Two Hundred Thirty Four Thousand Five Hundred Sixty Seven"

Example 4:

Input: num = 1234567891
Output: "One Billion Two Hundred Thirty Four Million Five Hundred Sixty Seven Thousand Eight Hundred Ninety One"

Constraints:

0 <= num <= 2^31 - 1

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，博主最开始的方法特别复杂，用了几个 switch 语句来列出所有的单词，但是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，首先说是要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，那么需要些一个处理三个一组数字的函数，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，然后还需要用些技巧，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;
    }
};