LeetCode 273. Integer to English Words

Posted 2020-10-14

此题是一个数字字符转化类问题。没有什么复杂的技巧，只要想清楚2个转化的规律：

1，此题给出的非负数字上限是2^31-1，大约是十亿的数量级(2^10 = 1024, (2^10)^3 ~1billion)。而我们已知每1000进位方式有三种后缀，thousand, million和billion,这就可以看作三重循环，刚好cover这个范围。

2.每重循环内部的命名规则是一致的，都是百位，十位和个位。唯一要注意的是在输出十位时，在20以内的命名都有专有名称，20以上都是*ten+个位。这几个情况并不复杂也不难实现（可分为 100位，大于20，小于20大于10，小于10大于0这四种情况）。

 1   vector<string> a ={"","Thousand ", "Million ", "Billion "};
 2     vector<string> b ={"Ten ", "Twenty ", "Thirty ", "Forty ", "Fifty ", "Sixty ", "Seventy ", "Eighty ", "Ninety "};
 3     vector<string> c ={"Ten ", "Eleven ", "Twelve ", "Thirteen ", "Fourteen ", "Fifteen ", "Sixteen ", "Seventeen ", "Eighteen ", "Nineteen "};
 4     vector<string> d ={"One ", "Two ", "Three ", "Four ", "Five ", "Six ", "Seven ", "Eight ", "Nine "};
 5     string numberToWords(int num) {
 6         string res;
 7         if(num == 0) return "Zero";
 8         for(int i=0; i<4; i++){
 9             if(num == 0) break;
10             int advance = num / 1000;
11             int left = num % 1000;
12             if(left > 0){
13                 res = convert(left) + a[i] + res; 
14             }
15             num = advance;
16         }
17         res.pop_back();
18         return res;
19     }
20     string convert(int num){
21         string res;
22         if(num >= 100){
23             res += d[num/100-1] + "Hundred ";
24             num = num % 100;
25         }
26         if(num >= 20){
27             res += b[num/10-1];
28             if(num % 10 > 0) res += d[num % 10 - 1];
29         }else if(num >= 10){
30             res += c[num - 10];
31         }else if(num > 0){
32             res += d[num-1];
33         }
34         return res;
35     }

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