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12. Integer to Roman/整数转罗马数字

12. Integer to Roman/整数转罗马数字

作者: 蜜糖_7474 | 来源:发表于2019-05-05 12:03 被阅读0次

    Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M.

    Symbol Value
    I 1
    V 5
    X 10
    L 50
    C 100
    D 500
    M 1000

    For example, two is written as II in 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: 3
    Output: "III"

    Example 2:

    Input: 4
    Output: "IV"

    Example 3:

    Input: 9
    Output: "IX"

    Example 4:

    Input: 58
    Output: "LVIII"
    Explanation: L = 50, V = 5, III = 3.

    Example 5:

    Input: 1994
    Output: "MCMXCIV"
    Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.

    AC代码

    class Solution {
    public:
        string intToRoman(int num) {
            string mp[4][11] = {
                {"", "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"}};
          //return mp[3][num / 1000] + mp[2][(num % 1000) / 100] +mp[1][(num % 100) / 10] + mp[0][num % 10];
            int i = 0, r;
            string ans;
            vector<string> vs;
            while (num != 0) {
                r = num % 10;
                vs.push_back(mp[i++][r]);
                num /= 10;
            }
            for (int i = vs.size() - 1; i >= 0; --i) ans += vs[i];
            return ans;
        }
    };
    

    总结

    有点像简单难度的题,注释那一条和下面的全部语句加起来效果一样

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