leetcode: Integer to Roman

最新推荐文章于 2022-12-16 00:28:08 发布

原创最新推荐文章于 2022-12-16 00:28:08 发布 · 701 阅读

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#Roman #C++ #leetcode

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本文介绍了一种将整数转换为罗马数字的算法，通过遍历字符数组并根据数值进行拼接，确保了从末位处理的便利性。特别注意4和9的特殊处理方式。

Given an integer, convert it to a roman numeral.

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

思路

同样，注意4和9的处理就行了。仍然是从末位处理比较方便。

class Solution {
public:
    string intToRoman(int num) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.    
		if(num<=0) return "";
		if(num>=4000) return "";
		int numb=num;
		char con[]={'I','V','X','L','C','D','M','\0'};
		char buff[16]={'\0'};
		int pos=0;
		int posb=0;
		while(numb!=0)
		{
			int digit=numb%10;
			if(digit<=3) while(digit-->0) buff[posb++]=con[pos*2];
			else if(digit==4){buff[posb++]=con[pos*2+1];buff[posb++]=con[pos*2];}
			else if(digit==5){buff[posb++]=con[pos*2+1];}
			else if(digit<=8){while(digit-->5) buff[posb++]=con[pos*2];buff[posb++]=con[pos*2+1];}
			else if(digit==9){buff[posb++]=con[pos*2+2];buff[posb++]=con[pos*2];}

			numb/=10;
			pos++;
		}
		int bLen=strlen(buff);
		for(int i=0;i<bLen/2;i++)
		{
			int tmp=buff[i];
			buff[i]=buff[bLen-i-1];
			buff[bLen-i-1]=tmp;
		}
        return string(buff);
    }
};