Leetcode13 Roman To Integer

easy的题，

import java.util.*;

public class romanToInteger13 {
    public int romanToInt(String s) {
        HashMap<String,Integer> map = new HashMap<>();
        int[] values =    {1000,900, 500,400,100, 90,  50, 40, 10,  9,   5,   4,  1};
        String[] romans = {"M","CM","D","CD","C","XC","L","XL","X","IX","V","IV","I"};
        for(int i=0;i<values.length;i++) map.put(romans[i], values[i]);
        int ans=0;
        for(int i=0;i<s.length();i++){
            if(i==s.length()-1) {ans+=map.get(s.substring(i));break;}
            if(map.containsKey(s.substring(i,i+2))) {
                ans+=map.get(s.substring(i,i+2));
                i++;
            }
            else {
                ans+=map.get(s.substring(i,i+1));
            }
        }
        return ans;
    }
}

43ms，90.62%。

下面是个33ms，100%的答案，思路相近，可以参考。

class Solution {
    public int value(char r){
        if (r == 'I') 
            return 1; 
        if (r == 'V') 
            return 5; 
        if (r == 'X') 
            return 10; 
        if (r == 'L') 
            return 50; 
        if (r == 'C') 
            return 100; 
        if (r == 'D') 
            return 500; 
        if (r == 'M') 
            return 1000; 
        return -1;     
    }
    public int romanToInt(String s) {
       int res=0;
        for(int i=0;i<s.length();i++){
          int s1=value(s.charAt(i));
          if(i+1<s.length()){
              int s2=value(s.charAt(i+1));
              if(s1>=s2){
                  res=res+s1;
              }
              else{
                  res=res+s2-s1;
                  i++;
              }
          }
            else{
                res = res + s1; 
                i++;  
            }
        }
        return res;
    }
}

它引入的side information更少。