Leetcode -- Day 9

最新推荐文章于 2024-10-10 17:10:04 发布

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原文链接：http://www.cnblogs.com/timoBlog/p/4653178.html

本文介绍了一种将罗马数字转换为整数的方法，并提供了两种不同的Java实现方案。第一种方案通过大量的条件判断来完成转换，而第二种方案则通过解析每个字符的价值，使得代码更加简洁易读。

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Leetcode -- Day 9

Question 1

Roman to Integer

Given a roman numeral, convert it to an integer.

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

This problem is not hard, But too many cases you need to consider, the code will be long to read.

http://www.rapidtables.com/convert/number/date-to-roman-numerals.htm

Got to this link to look some info. for Roman Number if you are not familiar with them like me.

Below is the initial one I thougt,

public class Solution {
    public int romanToInt(String s) {
        
        if (s==null || s.length()<1)
            return 0;
        s = s.trim();
        int i = 0;
        int result = 0;
        while (i < s.length())
        switch(s.charAt(i)){
            case 'I':
                i ++;
                if (i < s.length()){
                    switch(s.charAt(i)){
                        case 'V':
                            result += 4;
                            i ++;
                            break;
                        
                        case 'X':
                            result += 9;
                            i ++;
                            break;
                        default:
                            result += 1;
                            break;
                    }
                }
                else
                    result += 1;
                break;
                
            case 'V':
                i ++;
                result += 5;
                break;
            
            case 'X':
                i ++;
                if (i < s.length()){
                    switch(s.charAt(i)){
                        case 'L':
                            result += 40;
                            i ++;
                            break;
                         
                        case 'C':
                            result += 90;
                            i ++;
                            break;
                            
                        default:
                            result += 10;
                            break;
                    }
                }
                else
                    result += 10;
                break;
            
            case 'L':
                i ++;
                result += 50;
                break;
                
            case 'C':
                i ++;
                if (i < s.length()){
                    switch(s.charAt(i)){
                        case 'D':
                            result += 400;
                            i ++;
                            break;
                      
                        case 'M':
                            result += 900;
                            i ++;
                            break;
                            
                        default:
                            result += 100;
                            break;
                    }
                }
                else
                    result += 100;
                break;
            
            case 'D':
                i ++;
                result += 500;
                break;
                
            case 'M':
                i ++;
                result += 1000;
                break;
            
            default:
                return 0;
        }
        return result;
    }
}

Then I make a better one, seems a litter cleaner.

public class Solution {
    
    public int parseRomanNum(Character c){
        switch (c){
            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;
            default:
                return 0;
        }
    }
    
    public int romanToInt(String s) {
        
        if (s==null || s.length()<1)
            return 0;
        s = s.trim();
        
        int result = parseRomanNum(s.charAt(0));
        
        for (int i = 1; i < s.length(); i ++){
            int pre = parseRomanNum(s.charAt(i-1));
            int cur = parseRomanNum(s.charAt(i));
            if (pre >= cur)
                result += cur;
            else
                result = result - pre*2 + cur;
        }
        
        return result;
        
    }
    
}