leetcode Roman& Integer

最新推荐文章于 2018-12-25 22:43:35 发布

原创最新推荐文章于 2018-12-25 22:43:35 发布 · 237 阅读

0 ·

CC 4.0 BY-SA版权

算法专栏收录该内容

57 篇文章

订阅专栏

本文介绍了一种将整数转换为罗马数字以及罗马数字转换为整数的算法实现。包括通过遍历查找的方式进行整数到罗马数字的转换，以及通过比较左右数值大小来进行罗马数字到整数的转换。

参考：http://blog.youkuaiyun.com/ljiabin/article/details/39968583

1、罗马字母与数值的对应关系,罗马数字的范围0~3999

int toNumber(char ch) {
        switch (ch) {
            case 'I': return 1;
            case 'V': return 5;
            case 'X': return 10;
            case 'L': return 50;
            case 'C': return 100;
            case 'D': return 500;
            case 'M': return 1000;
        }
        return 0;
    }

2、CDL ==>550

转化方式：从右往左，如果左值比右值小，则减；如果左值比右值大，则加

public int romanToInt(String s) {
        char[] array=s.toCharArray();
        int result=toNumber(array[array.length-1]);
        for(int i=array.length-2;i>=0;i--){
            int t2=toNumber(array[i+1]);
            int t1=toNumber(array[i]);
            if(t1>=t2){
                result+=t1;
            }
            else{
                result-=t1;
            }
        }
        return result;
    }

3、转化方式：数值每个位置的数都表示一个罗马数字，如69（LXIX），其中LX和IX分别表示60和9.

roman的数组，表示{1,2,3，。。。10,20,30.。。。100,200,300.。。}

    public String intToRoman(int num) {
        String result="";
        String[][] roman={
            {"","I","II","III","IV","V","VI","VII","VIII","IX"},
            {"","X","XX","XXX","XL","L","LX","LXX","LXXX","XC"},
            {"","C","CC","CCC","CD","D","DC","DCC","DCCC","CM"},
            {"","M","MM","MMM"}
        };
        int col=0;
        while(num!=0){
            int remain=num%10;
            result=roman[col][remain]+result;
            col++;
            num=num/10;
        }
        return result;
    }

还有一种暴力方式：

class Solution {  
public:  
    string intToRoman(int num) {  
        string str;    
        string symbol[]={"M","CM","D","CD","C","XC","L","XL","X","IX","V","IV","I"};    
        int value[]=    {1000,900,500,400, 100, 90,  50, 40,  10, 9,   5,  4,   1};   
        for(int i=0;num!=0;++i)  
        {  
            while(num>=value[i])  
            {  
                num-=value[i];  
                str+=symbol[i];  
            }  
        }  
        return str;  
    }  
};