leetcode 13. 罗马数字转整数

正经解法

把一个小值放在大值的左边，就是做减法，否则为加法.

import java.util.*;

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;
        }
    }
}

替换法

比较取巧，好理解好实现，但是replace比较耗时

class Solution {
    public int romanToInt(String s) {
        s = s.replaceAll("IV", "IIII");
        s = s.replaceAll("IX", "VIIII");
        s = s.replaceAll("XL", "XXXX");
        s = s.replaceAll("XC", "LXXXX");
        s = s.replaceAll("CD", "CCCC");
        s = s.replaceAll("CM", "DCCCC");
        char[] chars = s.toCharArray();
        int ans = 0;
        for (char c : chars) {
            ans = ans + becomeInt(c);
        }
        return ans;
    }

    public int becomeInt(char c) {
        if (c == 'I') return 1;
        if (c == 'V') return 5;
        if (c == 'X') return 10;
        if (c == 'L') return 50;
        if (c == 'C') return 100;
        if (c == 'D') return 500;
        if (c == 'M') return 1000;
        return 0;
    }
}