[LintCode] Integer to Roman 整数转化成罗马数字

Given an integer, convert it to a roman numeral.

The number is guaranteed to be within the range from 1 to 3999.

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Clarification

What is Roman Numeral?

• https://en.wikipedia.org/wiki/Roman_numerals
• https://zh.wikipedia.org/wiki/%E7%BD%97%E9%A9%AC%E6%95%B0%E5%AD%97
• http://baike.baidu.com/view/42061.htm

Example

4 -> IV

12 -> XII

21 -> XXI

99 -> XCIX

more examples at: http://literacy.kent.edu/Minigrants/Cinci/romanchart.htm

LeetCode上的原题，请参见我之前的博客Integer to Roman。

解法一：

class Solution {
public:
    /**
     * @param n The integer
     * @return Roman representation
     */
    string intToRoman(int n) {
        string res = "";
        vector<vector<string>> v {{"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 cnt = ;
        for (int i = ; i >= ; --i) {
            int t = n / cnt;
            if (t) res += v[i][t - ];
            n %= cnt;
            cnt /= ;
        }
        return res;
    }
};

解法二：

class Solution {
public:
    /**
     * @param n The integer
     * @return Roman representation
     */
    string intToRoman(int n) {
        string res = "";
        vector<int> val {, , , , , , , , , , , , };
        vector<string> str{"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"};
        for (int i = ; i < val.size(); ++i) {
            while (n >= val[i]) {
                n -= val[i];
                res += str[i];
            }
        }
        return res;
    }
};

解法三：

class Solution {
public:
    /**
     * @param n The integer
     * @return Roman representation
     */
    string intToRoman(int n) {
        string res = "";
        vector<string> v1 {"", "I", "II", "III", "IV", "V", "VI", "VII", "VIII", "IX"};
        vector<string> v2 {"", "X", "XX", "XXX", "XL", "L", "LX", "LXX", "LXXX", "XC"};
        vector<string> v3 {"", "C", "CC", "CCC", "CD", "D", "DC", "DCC", "DCCC", "CM"};
        vector<string> v4 {"", "M", "MM", "MMM"};
        return v4[n / ] + v3[(n % ) / ] + v2[(n % ) / ] + v1[n % ];
    }
};