leetcode-【hard】273. Integer to English Words

题目：

273. Integer to English Words

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"

答案：

按着自己用英文阅读整数的顺序来就行：

1、这道题目不是难在思路上，而是难在考虑的情况比较多，比较多细节，一不小心就有小bug，需要思路清晰，逻辑清晰

2、英文中用来表示整数的单词并不多，分类：

1-9（个位数）,11-19（特殊的十位数），10-90（十位数），100（hundred），1000（thousand），1000000（million），1000000000（billion）

3、10这个十位数比较特别，只有在后两位完全为10时，才用ten

4、注意空格，后面没有数字了，就不能加空格了

5、每千位数跟后面的千位数（如果不为空，即0000000……）要有空格

6、注意不要乱加前缀空格和后缀空格

将上面的细节注意了就AC啦

代码：

 #include <vector>
 #include <string>

 using std::vector;
 using std::string;

 string n2s[] = {"One","Two","Three","Four","Five","Six","Seven","Eight","Nine"};
 string g2s[] = {"Eleven","Twelve","Thirteen","Fourteen","Fifteen","Sixteen","Seventeen","Eighteen","Nineteen"};
 string t2s[] = {"Ten","Twenty","Thirty","Forty","Fifty","Sixty","Seventy","Eighty","Ninety"};

 class Solution {
 private:
     vector<string> ans;

 public:
     string numberToWords(int num) {
         if(num == )
         {
             return "Zero";
         }

         unsigned int rem;
         unsigned int one;
         unsigned int two;
         unsigned int three;

         string perAns;
         ans.clear();

         do
         {
             //获取余数
             rem = num % ;
             num = num /;

             //存储每次计算的结果
             perAns.clear();
             perAns = "";

             //一千以内的数，取每一位
             one = rem % ;
             two = (rem / ) % ;
             three = (rem / ) % ;

             if(three != )
             {
                 perAns += n2s[three - ];
                 perAns += " Hundred";
             }

             //如果后两位都为0，那么后面就没有数了，也就不需要加空格
             if(perAns != "" && (one !=  || two != ))
             {
                 perAns += " ";
             }

             if(two == )
             {
                 //考虑最后一位是否为0的情况，因最后两位为10时，需要用“ten”
                 //否则就是十位数
                 if(one == )
                 {
                     perAns += t2s[];
                 }else
                 {
                     perAns += g2s[one - ];
                 }
             //考虑two不为0，不为1，则按一般规则去计算
             }else if(two != )
             {
                 perAns += t2s[two - ];

                 if(one != )
                 {
                     perAns += " ";//如果最后一位不为0，需要在其前面加空格
                     perAns += n2s[one - ];
                 }
             //考虑two为0的情况
             }else
             {
                 if(one != )
                 {
                     perAns += n2s[one - ];
                 }
             }

             //将结果存储到ans中，ans中的答案是以逆序形式存储了每个千位数
             ans.push_back(perAns);
         }while(num != );

         string result = "";
         unsigned int len = ans.size();
         //len最大长度为4，考虑每种位数就行，这里用的时候就会发现，合理使用goto，程序逻辑会很清晰
         switch(len)
         {
             case :goto three;break;
             case :goto two;break;
             case :goto one;break;
             case :goto zero;break;
         }

         three:
         if(ans[] != "")
         {
             result += ans[];
             result += " Billion";

             if(ans[] != "" || ans[] != "" || ans[] != "")
             {
                 result += " ";
             }
         }

         two:
         if(ans[] != "")
         {
             result += ans[];
             result += " Million";

             if(ans[] != "" || ans[] != "")
             {
                 result += " ";
             }
         }

         one:
         if(ans[] != "")
         {
             result += ans[];
             result += " Thousand";

             if(ans[] != "")
             {
                 result += " ";
             }
         }

         zero:
         if(ans[] != "")
         {
             result += ans[];
         }

         return result;
     }
 };

说明： 合理使用goto会取到很好的效果哦