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

Solution1:

 

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

 

　　这个用了hashmap，本质思想还是找到规律。如果当前字母大于后一个字母，加上当前字母的值；反之，减去当前字母的值。

solution2：

public int romanToInt(String s) {
    int nums[]=new int[s.length()];
    for(int i=0;i<s.length();i++){
        switch (s.charAt(i)){
            case 'M':
                nums[i]=1000;
                break;
            case 'D':
                nums[i]=500;
                break;
            case 'C':
                nums[i]=100;
                break;
            case 'L':
                nums[i]=50;
                break;
            case 'X' :
                nums[i]=10;
                break;
            case 'V':
                nums[i]=5;
                break;
            case 'I':
                nums[i]=1;
                break;
        }
    }
    int sum=0;
    for(int i=0;i<nums.length-1;i++){
        if(nums[i]<nums[i+1])
            sum-=nums[i];
        else
            sum+=nums[i];
    }
    return sum+nums[nums.length-1];
}

　　这个用了数组，和上面的方法差不多，更容易理解。