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：

  1 #include <vector>
  2 #include <string>
  3 
  4 using std::vector;
  5 using std::string;
  6 
  7 string n2s[] = {"One","Two","Three","Four","Five","Six","Seven","Eight","Nine"};
  8 string g2s[] = {"Eleven","Twelve","Thirteen","Fourteen","Fifteen","Sixteen","Seventeen","Eighteen","Nineteen"};
  9 string t2s[] = {"Ten","Twenty","Thirty","Forty","Fifty","Sixty","Seventy","Eighty","Ninety"};
 10 
 11 class Solution {
 12 private:
 13     vector<string> ans;
 14     
 15 public:
 16     string numberToWords(int num) {
 17         if(num == 0)
 18         {
 19             return "Zero";
 20         }
 21         
 22         unsigned int rem;
 23         unsigned int one;
 24         unsigned int two;
 25         unsigned int three;
 26         
 27         string perAns;
 28         ans.clear();
 29         
 30         do
 31         {
 32             //获取余数
 33             rem = num % 1000;
 34             num = num /1000;
 35             
 36             //存储每次计算的结果
 37             perAns.clear();
 38             perAns = "";
 39             
 40             //一千以内的数，取每一位
 41             one = rem % 10;
 42             two = (rem / 10) % 10;
 43             three = (rem / 100) % 10;
 44             
 45             if(three != 0)
 46             {
 47                 perAns += n2s[three - 1];
 48                 perAns += " Hundred";
 49             }
 50             
 51             //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
 52             if(perAns != "" && (one != 0 || two != 0))
 53             {
 54                 perAns += " ";
 55             }
 56             
 57             if(two == 1)
 58             {
 59                 //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”
 60                 //否则就是十位数
 61                 if(one == 0)
 62                 {
 63                     perAns += t2s[0];
 64                 }else
 65                 {
 66                     perAns += g2s[one - 1];
 67                 }
 68             //考虑two不为0，不为1，则按一般规则去计算
 69             }else if(two != 0)
 70             {
 71                 perAns += t2s[two - 1];
 72                 
 73                 if(one != 0)
 74                 {
 75                     perAns += " ";//如果最后一位不为0，需要在其前面加空格
 76                     perAns += n2s[one - 1];
 77                 }
 78             //考虑two为0的情况
 79             }else
 80             {
 81                 if(one != 0)
 82                 {
 83                     perAns += n2s[one - 1];
 84                 }
 85             }
 86             
 87             //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
 88             ans.push_back(perAns);
 89         }while(num != 0);
 90         
 91         string result = "";
 92         unsigned int len = ans.size();
 93         //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
 94         switch(len)
 95         {
 96             case 4:goto three;break;
 97             case 3:goto two;break;
 98             case 2:goto one;break;
 99             case 1:goto zero;break;
100         }
101         
102         three:
103         if(ans[3] != "")
104         {
105             result += ans[3];
106             result += " Billion";
107             
108             if(ans[2] != "" || ans[1] != "" || ans[0] != "")
109             {
110                 result += " ";
111             }
112         }
113         
114         two:
115         if(ans[2] != "")
116         {
117             result += ans[2];
118             result += " Million";
119             
120             if(ans[1] != "" || ans[0] != "")
121             {
122                 result += " ";
123             }
124         }
125         
126         one:
127         if(ans[1] != "")
128         {
129             result += ans[1];
130             result += " Thousand";
131             
132             if(ans[0] != "")
133             {
134                 result += " ";
135             }
136         }
137         
138         zero:
139         if(ans[0] != "")
140         {
141             result += ans[0];
142         }
143         
144         return result;
145     }
146 };

View Code

说明：合理使用goto会取到很好的效果哦

posted on 2016-07-16 13:17 Shirlies 阅读(379) 评论(0) 编辑收藏举报