1.问题描述

Given a roman numeral, convert it to an integer.Input is guaranteed to be within the range from 1 to 3999.

{"","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"}

题目翻译：
给定一个罗马数字，把这个罗马数字转换成整数，输入的数据范围1-3999

2.解题思路

这道题刚好和整数转罗马数字相反，那么就就可以利用相反的方法，把罗马数字所有的可能出现的情况建立一个数组，然后对各种情况进行判断，就能完成转换。在实现的时候，暴露出对java语言使用不熟悉的问题：
1.char和String的比较
2.substring()的范围问题
在实现这道题的时候边界情况也调整了好久，所以在下手写代码之前，一定要把所有的情况考虑清楚，不然调bug也很懵逼，为什么会出错也要想明白。
刚做这道题的时候，思路是一位一位的读取罗马字母，然后对特殊情况进行判断，如XC、CM，但是实现的时候对于特殊情况处理不好就放弃了，在上面的代码通过之后，我就看了排名第一的代码，也是这种思路，但是大神处理的很好，现在在代码中进行注释来说明。

3.代码实现

1.列举所有可能情况，进行解析

class Solution {
    public int romanToInt(String s) {
		int len = s.length();
		int res = 0;
		if(len==0||s==null)
			res = 0;
		int j=0;
		String[] arr= {"M","CM","D","CD","C","XC","L","XL","X","IX","V","IV","I"};
        //数组保存所有情况
		int [] num = {1000,900,500,400,100,90,50,40,10,9,5,4,1};
		for(int i =0;i<len;) {
            //循环用于处理重复出现的字符
			while(i+1<=len&&arr[j].equals(s.substring(i,i+1))) {
				res+=num[j];
				i++;
			}	
            //判断避免越界	
			if(i+2<=len) { 
                //判断包含两个字符的情况
                if(arr[j].equals(s.substring(i, i+2)) ) {
                    res+=num[j];
                    j++;
                    i = i+2;
                    continue;
                }
			}
            //防止越界，同时跳过不匹配的情况
			if(i+1<=len && !arr[j].equals(s.substring(i,i+1))) {
				j++;
				continue;
			}
	
			
			
			}
		return res;
    }
}

2.逐位解析，处理特殊情况

class Solution {
    public int romanToInt(String s) {
         int[] chs = new int[s.length()];
        int result = 0;
        //把多有字符代表的数值列举出来，一位一位的对应，
        //但是这里存在类似CD和DC逆序的问题
        for (int i = 0; i < chs.length; i++) {
            switch (s.charAt(i)) {
                case 'I':
                    chs[i] = 1; break;
                case 'V':
                    chs[i] = 5; break;
                case 'X':
                    chs[i] = 10; break;
                case 'L':
                    chs[i] = 50; break;
                case 'C':
                    chs[i] = 100; break;
                case 'D':
                    chs[i] = 500; break;
                case 'M':
                    chs[i] = 1000; break;
            }
        }
        for (int i = 0; i < chs.length; i++){
            if ( i == chs.length - 1){
                result += chs[i];
            } 
            else if (chs[i] < chs [i + 1]){
                //处理顺序不同的情况
                result = result - chs [i] ;
            } else 
                result += chs[i];

        }
        return result;
    }
}