【一天一道LeetCode】#12 Integer to Roman

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一天一道LeetCode系列

（一）题目

Given an integer, convert it to a roman numeral.

Input is guaranteed to be within the range from 1 to 3999.

（二）解题

将整形数字转换成罗马数字
罗马数字共有七个，即I(1)，V(5)，X(10)，L(50)，C(100)，D(500)，M(1000)
举例：Ⅰ、Ⅱ、Ⅲ、Ⅳ、Ⅴ、Ⅵ、Ⅶ、Ⅷ、Ⅸ、Ⅹ、Ⅺ、Ⅻ表示1-11
代码：

class Solution {public:    string intToRoman(int num) {        string str = "";        string roman[8] = {"I","V","X","L","C","D","M"};        int j=0;        while(num)        {            int temp = num%10;            string str_t ="";            if(temp <4)            {                int count = temp;                while(count--) str_t+=roman[j];            }            else if(temp==4)            {                str_t=roman[j];                str_t+=roman[j+1];            }            else if(temp>=5&&temp<=8)            {                str_t+=roman[j+1];                int count = temp-5;                while(count--) str_t+=roman[j];            }            else if(temp==9)            {                str_t+=roman[j];                str_t+=roman[j+2];            }            cout<<roman[j]<<endl;            cout<<str_t<<endl;            str =str_t+str;            j=j+2;            num = num/10;        }        return str;    }};

我这段代码看起来比较臃肿。看到网上的写法如beiyeqingteng博主的：

public class Solution {      public String intToRoman(int number) {          int[] values = {1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1 };          String[] numerals = {"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I" };          StringBuilder result = new StringBuilder();          for (int i = 0; i < values.length; i++) {              while (number >= values[i]) {                  number -= values[i];                  result.append(numerals[i]);              }          }          return result.toString();      }  }

看起来就简单多了。

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