Type：medium
Roman numerals are represented by seven different symbols:I, V, X, L, C, D and M.
SymbolValueI 1V 5X 10L 50C 100D 500M 1000
For example, two is written as IIin Roman numeral, just two one's added together. Twelve is written as, XII, which is simply X + II. The number twenty seven is written as XXVII, which is XX + V + II.
Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not IIII. Instead, the number four is written as IV. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as IX. There are six instances where subtraction is used:
I can be placed before V (5) and X (10) to make 4 and 9.
X can be placed before L (50) and C (100) to make 40 and 90.
C can be placed before D (500) and M (1000) to make 400 and 900.
Given an integer, convert it to a roman numeral. Input is guaranteed to be within the range from 1 to 3999.
Example 1:
Input:3Output:"III"
Example 2:
Input:4Output:"IV"
Example 3:
Input:9Output:"IX"
Example 4:
Input:58Output:"LVIII"Explanation:L = 50, V = 5, III = 3.
Example 5:
Input:1994Output:"MCMXCIV"Explanation:M = 1000, CM = 900, XC = 90 and IV = 4
本题与13题相反，是将int型转为罗马数字。这道题真的孤儿，如果测试题出这类型绝对脑子有洞。
我的想法很简单，读入一个数值后，从个位到千位依次读取并转换为罗马数字。因此首先，将代表1-9,10-90，100-900,1000-3000的30种数值表示方法写进字符数组。之后挨个读取并写进结果中。
本题的考点：
1、如何定义字符数组。
2、返回的ret = s[count][temp] + ret，而不是ret += s[count][temp]。即数值大的罗马数字在前。
class Solution {
public:
string intToRoman(int num) {
string s[4][10] = {{"", "I", "II", "III", "IV", "V", "VI", "VII", "VIII", "IX"},
{"", "X", "XX", "XXX", "XL", "L", "LX", "LXX", "LXXX", "XC"},
{"", "C", "CC", "CCC", "CD", "D", "DC", "DCC", "DCCC", "CM"},
{"", "M", "MM", "MMM"}};
int count = 0;
string ret;
while(num != 0){
int temp = num % 10;
ret = s[count][temp] + ret;
num = num / 10;
count++;
}
return ret;
}
};
