leetcode解题: Roman to Integer (13)

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

解法1：O(N) Time

Roman to Integer 是较容易的。需要记忆的是几个关键的Roman字母和数值的对应关系。
从后往前扫描，如果前面的数值比后面的数值小，则需要减去，否则加上。

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| I | V | X  | L | C | D | M  |
|---|---|----|---|---|---|--- |
| 1 | 5 | 10 |50 |100|500|1000|

C++

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class Solution {
  public:
  int romanToInt(string s) {
      if (s.empty()) {
          return 0;
      }
      int res = 0;
      res += romans[toupper(s[s.size() - 1])];
      for (int i = s.size() - 2; i >= 0; --i) {
          if (romans[toupper(s[i])] < romans[toupper(s[i + 1])]) {
              res -= romans[toupper(s[i])];
          } else {
              res += romans[toupper(s[i])];
          }
      }
      return res;
  }
  private:
    unordered_map<char, int> romans {{'I',1},{'V',5},{'X',10}, {'L',50}, {'C', 100}, {'D',500}, {'M',1000}};
}
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class Solution {
    public int romanToInt(String s) {
        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);
        int res = 0;
        res += map.get(s.charAt(s.length() - 1));
        for (int i = s.length() - 2; i >= 0; i--) {
            int current = map.get(s.charAt(i));
            int last = map.get(s.charAt(i + 1));
            if (current < last) {
                res -= current;
            } else {
                res += current;                
            }
        }
        return res;        
    }
}