leetcode 13. Roman to Integer（罗马字母转数字）

Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M.

Symbol       Value
I             1
V             5
X             10
L             50
C             100
D             500
M             1000

For example, 2 is written as II in Roman numeral, just two ones added together. 12 is written as XII, which is simply X + II. The number 27 is written as XXVII, which is XX + V + II.

Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not IIII. Instead, the number four is written as IV. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as IX. There are six instances where subtraction is used:

I can be placed before V (5) and X (10) to make 4 and 9.
X can be placed before L (50) and C (100) to make 40 and 90.
C can be placed before D (500) and M (1000) to make 400 and 900.
Given a roman numeral, convert it to an integer.

Example 1:

Input: s = “III”
Output: 3
Explanation: III = 3.
Example 2:

Input: s = “LVIII”
Output: 58
Explanation: L = 50, V= 5, III = 3.

罗马字母里面有7个字母对应相应的数字，
然后有几个字母的两两组合：
LV, LX, XL, XC, CD, CM单独对应了数字。
把指定的罗马字母转为数字。

思路：

确定好一一对应关系即可，
遍历s。
因为只有几个字母，switch-case即可。

这里遇到‘I’ 是无条件+1的，那么遇到’X’后如果它前面是’I’,
‘IX’本身是9，而前面’I’ 已经加了1，这时只需再+8。
同理其他相减的部分。

public int romanToInt(String s) {
    int num = 0;
    
    for(int i = 0; i < s.length(); i ++) {
        switch(s.charAt(i)) {
            case 'I':
                num += 1; break;
            case 'V':
                if(i > 0 && s.charAt(i-1) == 'I') num +=3;
                else num += 5;
                break;
            case 'X':
                if(i > 0 && s.charAt(i-1) == 'I') num +=8;
                else num += 10;
                break;
            case 'L':
                if(i > 0 && s.charAt(i-1) == 'X') num +=30;
                else num += 50;
                break;
            case 'C':
                if(i > 0 && s.charAt(i-1) == 'X') num +=80;
                else num += 100;
                break;
            case 'D':
                if(i > 0 && s.charAt(i-1) == 'C') num +=300;
                else num += 500;
                break;
            case 'M':
                if(i > 0 && s.charAt(i-1) == 'C') num +=800;
                else num += 1000;
                break;
        }
    }
    
    return num;
}