LeetCode 13 Roman to Integer

题目：

Roman numerals are represented by seven different symbols: I, V, X, L, C, Dand M.

Symbol       Value
I             1
V             5
X             10
L             50
C             100
D             500
M             1000

For example, two is written as II in Roman numeral, just two one's added together. Twelve is written as, XII, which is simply X + II. The number twenty seven is written as XXVII, which is XX + V + II.

Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not IIII. Instead, the number four is written as IV. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as IX. There are six instances where subtraction is used:

• I can be placed before V (5) and X (10) to make 4 and 9. 
• X can be placed before L (50) and C (100) to make 40 and 90. 
• C can be placed before D (500) and M (1000) to make 400 and 900.

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

Example 1:

Input: "III"
Output: 3

Example 2:

Input: "IV"
Output: 4

Example 3:

Input: "IX"
Output: 9

Example 4:

Input: "LVIII"
Output: 58
Explanation: L = 50, V= 5, III = 3.

Example 5:

Input: "MCMXCIV"
Output: 1994
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.

解法：

可以看出一般情况下，字符代表的数字越大则越靠前，而特殊情况即代表数字小的字符出现在代表数字大的字符前时，则需减去代表数字小的字符代表的数字。故可以遍历字符串，若当前字符代表的数字的与后面一个字符代表的数字相比较大时，则在结果加上当前字符代表的数字，否则减去当前字符代表的数字，最后一位单独判断，以免造成数组访问越界。

为了更好的比较字符代表的数字的大小关系，使用hashmap存储对应关系。

class Solution {
    public int romanToInt(String s) {
        
        int res = 0;
        Map<Character,Integer> map = new HashMap<>();
        
        map.put('I',1);
        map.put('V',5);
        map.put('X',10);
        map.put('L',50);
        map.put('C',100);
        map.put('D',500);
        map.put('M',1000);
        
        for(int i=0;i<s.length()-1;i++)
        {
            if(map.get(s.charAt(i))<map.get(s.charAt(i+1)))
                res -= map.get(s.charAt(i));
            else
                res += map.get(s.charAt(i));
        }
        res += map.get(s.charAt(s.length()-1));
        
        return res;
    }
}