【LeetCode】罗马数字转整数

一时失言乱红尘 2021-11-04 16:08 524阅读 0赞

Java code

class Solution {
    public int romanToInt(String s) {
        int sum = 0;
        char[] c = s.toCharArray();
        for(int i = 0; i < c.length; i++) {
            if(c[i] == 'I') {
                if(i != (c.length - 1) && c[i + 1] == 'V') {
                    sum += 4;
                    i++;
                    continue;
                }
                if(i != (c.length - 1) && c[i + 1] == 'X') {
                    sum += 9;
                    i++;
                    continue;
                }
                 sum += 1;
            }    
            if(c[i] == 'V') {
                sum += 5;
            }
            if(c[i] == 'X') {
                if(i != (c.length - 1) && c[i + 1] == 'L') {
                    sum += 40;
                    i++;
                    continue;
                }
                if(i != (c.length - 1) && c[i + 1] == 'C') {
                    sum += 90;
                    i++;
                    continue;
                }
                sum += 10;
            }
            if(c[i] == 'L') {
                sum += 50;
            }
            if(c[i] == 'C') {
                if(i != (c.length - 1) && c[i + 1] == 'D') {
                    sum += 400;
                    i++;
                    continue;
                }
                if(i != (c.length - 1) && c[i + 1] == 'M') {
                    sum += 900;
                    i++;
                    continue;
                }
                sum += 100;
            }
            if(c[i] == 'D') {
                sum += 500;
            }
            if(c[i] == 'M') {
                sum += 1000;
            }
        }
        return sum;
    }
}