罗马数字转整数

HashMap解法9ms,40.2 MB

class Solution {
    public int romanToInt(String s) {
        HashMap<String,Integer> map = new HashMap<>();
        map.put("M",1000);
        map.put("CM",900);
        map.put("D",500);
        map.put("CD",400);
        map.put("C",100);
        map.put("XC",90);
        map.put("L",50);
        map.put("XL",40);
        map.put("X",10);
        map.put("IX",9);
        map.put("V",5);
        map.put("IV",4);
        map.put("I",1);

        int ans = 0;
        for(int i = 0;i<s.length();){
            if(i+1<s.length()&&map.containsKey(s.substring(i,i+2))){
                ans += map.get(s.substring(i,i+2));
                i += 2;
            }else{
                ans += map.get(s.substring(i,i+1));
                i++;
            }
        }
        return ans;
    }
}

switch解法3ms,39.8 MB

class Solution {
    public int romanToInt(String s) {
        int sum = 0;
        int preNum = getValue(s.charAt(0));
        for(int i = 1;i < s.length(); i ++) {
            int num = getValue(s.charAt(i));
            if(preNum < num) {
                sum -= preNum;
            } else {
                sum += preNum;
            }
            preNum = num;
        }
        sum += preNum;
        return sum;
    }
    
    private int getValue(char ch) {
        switch(ch) {
            case 'I': return 1;
            case 'V': return 5;
            case 'X': return 10;
            case 'L': return 50;
            case 'C': return 100;
            case 'D': return 500;
            case 'M': return 1000;
            default: return 0;
        }
    }
}