LeetCode 12 — Integer to Roman(整數轉羅馬數字)

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: C = 100, L = 50, XXX = 30 and III = 3.
Example 5:
Input: 1994
Output: “MCMXCIV”
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.

翻譯
羅馬數字包含以下七種字符： I， V， X， L，C，D 和 M。
字符 數值
I 1
V 5
X 10
L 50
C 100
D 500
M 1000
例如， 羅馬數字 2 寫做 II ，即爲兩個並列的 1。12 寫做 XII ，即爲 X + II 。 27 寫做 XXVII, 即爲 XX + V + II 。
通常情況下，羅馬數字中小的數字在大的數字的右邊。但也存在特例，例如 4 不寫做 IIII，而是 IV。數字 1 在數字 5 的左邊，所表示的數等於大數 5 減小數 1 得到的數值 4 。同樣地，數字 9 表示爲 IX。這個特殊的規則只適用於以下六種情況：
I 可以放在 V (5) 和 X (10) 的左邊，來表示 4 和 9。
X 可以放在 L (50) 和 C (100) 的左邊，來表示 40 和 90。
C 可以放在 D (500) 和 M (1000) 的左邊，來表示 400 和 900。
給定一個整數，將其轉爲羅馬數字。輸入確保在 1 到 3999 的範圍內。

示例 1:
輸入: 3
輸出: “III”
示例 2:
輸入: 4
輸出: “IV”
示例 3:
輸入: 9
輸出: “IX”
示例 4:
輸入: 58
輸出: “LVIII”
解釋: C = 100, L = 50, XXX = 30, III = 3.
示例 5:
輸入: 1994
輸出: “MCMXCIV”
解釋: M = 1000, CM = 900, XC = 90, IV = 4.

分析
千，百，十，個位分開討論即可。

c++實現

class Solution {
public:
    string intToRoman(int num) {
        int qian = num/1000;
        int bai = num/100%10;
        int shi = num/10%10;
        int ge = num%10;
        string s = "";
        for (int i = 1; i <= qian; i++)
            s += "M";
        
        if (bai == 4)
            s += "CD";
        if (bai == 9)
            s += "CM";
        if (bai < 4)
        {
            for (int i = 1; i <= bai; i++)
                s += "C";
        }
        if (bai > 4 && bai < 9)
        {
            s += "D";
            for (int i = 6; i <= bai; i++)
                s += "C";
        }
        
        if (shi == 4)
            s += "XL";
        if (shi == 9)
            s += "XC";
        if (shi < 4)
        {
            for (int i = 1; i <= shi; i++)
                s += "X";
        }
        if (shi > 4 && shi < 9)
        {
            s += "L";
            for (int i = 6; i <= shi; i++)
                s += "X";
        }
        
        if (ge == 4)
            s += "IV";
        if (ge == 9)
            s += "IX";
        if (ge < 4)
        {
            for (int i = 1; i <= ge; i++)
                s += "I";
        }
        if (ge > 4 && ge < 9)
        {
            s += "V";
            for (int i = 6; i <= ge; i++)
                s += "I";
        }
        return s;
    }
};