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）。

Corner Case: 1) num =0时； 2)如果是倒序生成，需注意最后次序；3) 1,000,000情况，千位无需表达； 4)空格处理，在每个字符后先加空格

1   vector
      
        a ={
    "","Thousand ", "Million ", "Billion "}; 2     vector
       
         b ={
    "Ten ", "Twenty ", "Thirty ", "Forty ", "Fifty ", "Sixty ", "Seventy ", "Eighty ", "Ninety "}; 3     vector
        
          c ={
    "Ten ", "Eleven ", "Twelve ", "Thirteen ", "Fourteen ", "Fifteen ", "Sixteen ", "Seventeen ", "Eighteen ", "Nineteen "}; 4     vector
         
           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     }