[LeetCode]12. Integer to Roman

最新推荐文章于 2022-10-20 18:00:00 发布

Winter_ann

最新推荐文章于 2022-10-20 18:00:00 发布

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本文链接：https://blog.youkuaiyun.com/Winter_ann/article/details/50812535

本文介绍了一种将整数转换为罗马数字的算法实现，适用于1到3999范围内的整数。通过逐位处理的方式，使用switch-case结构匹配每一位数字对应的罗马数字表示。

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Given an integer, convert it to a roman numeral.

Input is guaranteed to be within the range from 1 to 3999.

思路：分别计算每一位上的数字的表示形式，最后连接字符串

   public String intToRoman(int num) {
		int a = num%10;
		String s1 = "";
		switch(a) {
		case 1 : s1 = "I"; break;
		case 2 : s1 = "II"; break;
		case 3 : s1 = "III";break;
		case 4 : s1 = "IV"; break;
		case 5 : s1 = "V"; break;
		case 6 : s1 = "VI"; break;
		case 7 : s1 = "VII"; break;
		case 8 : s1 = "VIII"; break;
		case 9 : s1 = "IX"; break;
		case 0 :  break;
		}
		if(num/10 == 0) return s1;
		else {
			String s2 = "";
			int b = (num/10)%10;
			switch(b) {
			case 1 : s2 = "X"; break;
			case 2 : s2 = "XX"; break;
			case 3 : s2 = "XXX";break;
			case 4 : s2 = "XL"; break;
			case 5 : s2 = "L"; break;
			case 6 : s2 = "LX"; break;
			case 7 : s2 = "LXX"; break;
			case 8 : s2 = "LXXX"; break;
			case 9 : s2 = "XC"; break;
			case 0 :  break;
			}
			if(num/100 == 0) return s2+s1; 
			else {
				String s3 = "";
				int c = (num/100)%10;
				switch(c) {
				case 1 : s3 = "C"; break;
				case 2 : s3 = "CC";break;
				case 3 : s3 = "CCC"; break;
				case 4 : s3 = "CD"; break;
				case 5 : s3 = "D"; break;
				case 6 : s3 = "DC"; break;
				case 7 : s3 = "DCC"; break;
				case 8 : s3 = "DCCC"; break;
				case 9 : s3 = "CM"; break;
				case 0 :  break;
				}
				if(num/1000 == 0) return s3+s2+s1; 
				else {
					String s4 = "";
					int d = (num/1000)%10;
					switch(d) {
					case 1 : s4 = "M"; break;
					case 2 : s4 = "MM";break;
					case 3 : s4 = "MMM"; break;
					}
					return  s4+s3+s2+s1;
				}
			}  
		}

	}