【LeetCode】12. Integer to Roman（C++）

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本文详细解析了一道LeetCode上的经典算法题——将整数转换为罗马数字。通过两种不同的实现方式，一种是逐个判断并减去相应数值，另一种是使用数组优化，最后介绍了一种更简洁的方法，通过分解千、百、十、个位来快速生成罗马数字。文章包含了完整的代码实现，是学习和理解罗马数字转换规则的绝佳资源。

地址：https://leetcode.com/problems/integer-to-roman/

题目：

Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M.
在这里插入图片描述
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.

理解：

似乎感觉刷过roman转integer的问题？翻了下没找到。
嗯。。愚蠢的我写出了下面的实现=_=

实现：

class Solution {
public:
	string intToRoman(int num) {
		string res;
		while (num) {
			if (num >= 1000) {
				res += "M";
				num -= 1000;
			}
			else if (num >= 900) {
				res += "CM";
				num -= 900;
			}
			else if (num >= 500) {
				res += "D";
				num -= 500;
			}
			else if (num >= 400) {
				res += "CD";
				num -= 400;
			}
			else if (num >= 100) {
				res += "C";
				num -= 100;
			}
			else if (num >= 90) {
				res += "XC";
				num -= 90;
			}
			else if (num >= 50) {
				res += "L";
				num -= 50;
			}
			else if (num >= 40) {
				res += "XL";
				num -= 40;
			}
			else if (num >= 10) {
				res += "X";
				num -= 10;
			}
			else if (num >= 9) {
				res += "IX";
				num -= 9;
			}
			else if (num >= 5) {
				res += "V";
				num -= 5;
			}
			else if (num >= 4) {
				res += "IV";
				num -= 4;
			}
			else {
				res += "I";
				num--;
			}
		}
		return res;
	}
};

哈哈哈虽然很罗嗦，总是为数不多的一次就ac的题，，其实最简单的优化就是用两个数组，不用这么搞了

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;
    }
};

因为这个问题里的数字是有范围的，因此可以根据这一个特点进行简化。手动算出千，百，十，个位的数字，并按照规则加在一起。

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