Question

link

Stats

Frequency	4
Diffficulty	2
Adjusted Difficulty	2
Time to use	----------

Ratings/Color = 1(white) 2(lime) 3(yellow) 4/5(red)

Analysis

This question is easier than [LeetCode 12] Integer to Roman.

The basic idea is to read 2 char (as pre and cur) and then check. If pre > cur, then OK, just do the addition. If pre < cur, we need to subtract (2 * pre) from the result.

The code is easy and concise if you have the idea.

Solution

The code explains itself.

My code

public class Solution {
    public int romanToInt(String s) {
        if (s == null || s.length() == 0) {
            return 0;
        }
        int ans = 0;
        int pre = 0;
        int cur = 0;
        for (int i = 0; i < s.length(); i++) {
            cur = getNum(s.charAt(i));
            if (i == 0 || pre >= cur) {
                ans += cur;
            } else {
                ans += cur - (2 * pre);
            }
            pre = cur;
        }
        return ans;
    }

    private int getNum(char a){
        switch(a){
            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;
        }
        return 0;
    }
}

We can also do only adding. Someone posted the code in here. Read it ONLY if you are interested.

class Solution {
public:
    int romanToInt(string s) {
        // 4:IV, 9:IX, 40:XL, 90:XC, 400:CD, 900:CM,
        // 1:I, 10:X, 100:C, 1000:M
        int res=0;
        char pre = ' ';
        for(int i=0;i<s.size();i++){
            if (s[i]=='M' && pre!='C') {res+=1000;}
            if (s[i]=='C' && pre!='X') {res+=100;}
            if (s[i]=='X' && pre!='I') {res+=10;}

            if (s[i]=='M' && pre=='C') {res+=800;}
            if (s[i]=='C' && pre=='X') {res+=80;}
            if (s[i]=='X' && pre=='I') {res+=8;}

            if (s[i]=='I' ) {res+=1;}

            if (s[i]=='V' && pre!='I'){res+=5;}
            if (s[i]=='L' && pre!='X'){res+=50;}
            if (s[i]=='D' && pre!='C'){res+=500;}

            if (s[i]=='V' && pre=='I'){res+=3;}
            if (s[i]=='L' && pre=='X'){res+=30;}
            if (s[i]=='D' && pre=='C'){res+=300;}

            pre = s[i];
        }
        return res;
    }
};