leetcode 012 整数转罗马数字

最新推荐文章于 2025-08-12 12:49:47 发布

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最新推荐文章于 2025-08-12 12:49:47 发布

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CC 4.0 BY-SA版权

文章标签： leetcode

本文链接：https://blog.youkuaiyun.com/weixin_43330995/article/details/88046505

本文介绍了一种将整数转换为罗马数字的算法实现，通过分解整数为千、百、十、个位，并针对每位数的不同情况，转换为对应的罗马数字表示。此方法虽然简单直观，但效率较低。

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很傻的方法，穷举；
用4个int的数组，分别存放num的千、百、十、个位，然后针对每一位列出各种情况；

事实证明，占用内存和耗时都很高；

class Solution {
public:
	string intToRoman(int num) 
	{
		int temp[4] = {0};
		string res;
		for (int i = 0; i < 4; i++)
		{
			temp[3-i] = num % 10;
			num /= 10;
		}
		for (int i = 0; i < temp[0]; i++)
		{
			res += "M";
		}
		switch (temp[1])
		{
		case 0:break;
		case 1:res += "C"; break;
		case 2:res += "C"; res += "C"; break;
		case 3:res += "C"; res += "C"; res += "C"; break;
		case 4:res += "CD"; break;
		case 5:res += "D"; break;
		case 6:res += "D"; res += "C"; break;
		case 7:res += "D"; res += "C"; res += "C"; break;
		case 8:res += "D"; res += "C"; res += "C"; res += "C"; break;
		case 9:res += "CM"; break;
		}
		switch (temp[2])
		{
		case 0:break;
		case 1:res += "X"; break;
		case 2:res += "X"; res += "X"; break;
		case 3:res += "X"; res += "X"; res += "X"; break;
		case 4:res += "XL"; break;
		case 5:res += "L"; break;
		case 6:res += "L"; res += "X"; break;
		case 7:res += "L"; res += "X"; res += "X"; break;
		case 8:res += "L"; res += "X"; res += "X"; res += "X"; break;
		case 9:res += "XC"; break;
		}
		switch (temp[3])
		{
		case 0:break;
		case 1:res += "I"; break;
		case 2:res += "I"; res += "I"; break;
		case 3:res += "I"; res += "I"; res += "I"; break;
		case 4:res += "IV"; break;
		case 5:res += "V"; break;
		case 6:res += "V"; res += "I"; break;
		case 7:res += "V"; res += "I"; res += "I"; break;
		case 8:res += "V"; res += "I"; res += "I"; res += "I"; break;
		case 9:res += "IX"; break;
		}
		return res;
	}
};