leetcode 012 整数转罗马数字

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。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

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;
	}
};
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值