C++实现LeetCode(12.整数转化成罗马数字)

[LeetCode] 12. Integer to Roman 整数转化成罗马数字

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: L = 50, V = 5, III = 3.

Example 5:

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

之前那篇文章写的是罗马数字转化成整数 Roman to Integer， 这次变成了整数转化成罗马数字，基本算法还是一样。由于题目中限定了输入数字的范围 (1 - 3999), 使得题目变得简单了不少。

I - 1

V - 5

X - 10

L - 50

C - 100 

D - 500

M - 1000

例如整数 1437 的罗马数字为 MCDXXXVII， 我们不难发现，千位，百位，十位和个位上的数分别用罗马数字表示了。 1000 - M, 400 - CD, 30 - XXX, 7 - VII。所以我们要做的就是用取商法分别提取各个位上的数字，然后分别表示出来：

100 - C

200 - CC

300 - CCC

400 - CD

500 - D

600 - DC

700 - DCC

800 - DCCC

900 - CM

可以分为四类，100 到 300 一类，400 一类，500 到 800 一类，900 最后一类。每一位上的情况都是类似的，代码如下：

解法一：

class Solution {
public:
    string intToRoman(int num) {
        string res = "";
        vector<char> roman{'M', 'D', 'C', 'L', 'X', 'V', 'I'};
        vector<int> value{1000, 500, 100, 50, 10, 5, 1};
        for (int n = 0; n < 7; n += 2) {
            int x = num / value[n];
            if (x < 4) {
                for (int i = 1; i <= x; ++i) res += roman[n];
            } else if (x == 4) {
                res = res + roman[n] + roman[n - 1]; 
            } else if (x > 4 && x < 9) {
                res += roman[n - 1];
                for (int i = 6; i <= x; ++i) res += roman[n];
            } else if (x == 9) {
                res = res + roman[n] + roman[n - 2];
            }
            num %= value[n];            
        }
        return res;
    }
};

本题由于限制了输入数字范围这一特殊性，故而还有一种利用贪婪算法的解法，建立一个数表，每次通过查表找出当前最大的数，减去再继续查表，参见代码如下：

解法二：

class Solution {
public:
    string intToRoman(int num) {
        string res = "";
        vector<int> val{1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1};
        vector<string> str{"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"};
        for (int i = 0; i < val.size(); ++i) {
            while (num >= val[i]) {
                num -= val[i];
                res += str[i];
            }
        }
        return res;
    }
};

下面这种方法个人感觉属于比较投机取巧的方法，把所有的情况都列了出来，然后直接按位查表，O(1) 的时间复杂度啊，参见代码如下：

解法三：

class Solution {
public:
    string intToRoman(int num) {
        string res = "";
        vector<string> v1{"", "M", "MM", "MMM"};
        vector<string> v2{"", "C", "CC", "CCC", "CD", "D", "DC", "DCC", "DCCC", "CM"};
        vector<string> v3{"", "X", "XX", "XXX", "XL", "L", "LX", "LXX", "LXXX", "XC"};
        vector<string> v4{"", "I", "II", "III", "IV", "V", "VI", "VII", "VIII", "IX"};
        return v1[num / 1000] + v2[(num % 1000) / 100] + v3[(num % 100) / 10] + v4[num % 10];
    }
};