| 标题： | Integer to Roman |
| 通过率： | 34.4% |
| 难度： | 中等 |
Given an integer, convert it to a roman numeral.
Input is guaranteed to be within the range from 1 to 3999.
前面做过数字转罗马数字，相对来说比较简单，从后面向前面走。本题是数字转罗马就是先看有几个1000，num/1000在看有几个一百，（num%1000）/100,在看有几个十，（num%100）/10，在看有几个个位数，num%10，然后对应的去查找数字表：
map={{"0","I","II","III","IV","V","VI","VII","VIII","IX"},{"0","X","XX","XXX","XL","L","LX"
,"LXX","LXXX","XC"},{"0","C","CC","CCC","CD","D",
"DC","DCC","DCCC","CM"},{"0","M","MM","MMM"}};
因为没有零，所以二维数组是从位置1开始的，具体看代码：
1 public class Solution { 2 public String intToRoman(int num) { 3 String [][] map={{"0","I","II","III","IV","V","VI","VII","VIII","IX"},{"0","X","XX","XXX","XL","L","LX" 4 ,"LXX","LXXX","XC"},{"0","C","CC","CCC","CD","D", 5 "DC","DCC","DCCC","CM"},{"0","M","MM","MMM"}}; 6 String result=""; 7 if(num/1000!=0)result+=map[3][num/1000]; 8 if((num%1000)/100!=0)result+=map[2][(num%1000)/100]; 9 if((num%100)/10!=0)result+=map[1][(num%100)/10]; 10 if(num%10!=0)result+=map[0][num%10]; 11 return result; 12 13 } 14 }
python代码：
1 class Solution: 2 def intToRoman(self, num): 3 numeral_map = {1: "I", 4: "IV", 5: "V", 9: "IX", 10: "X", 40: "XL", 50: "L", 90: "XC", 100: "C", 400: "CD", 500: "D", 900: "CM", 1000: "M"} 4 keyset, length, result = sorted(numeral_map.keys()), len(numeral_map), "" 5 while num > 0: 6 for i in reversed(range(length)): 7 key = keyset[i] 8 while num / key > 0: 9 result += numeral_map[key] 10 num -= key 11 return result