LeetCode 13. Roman to Integer

原题链接在这里：https://leetcode.com/problems/roman-to-integer/

题目：

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, two is written as II in Roman numeral, just two one's added together. Twelve is written as, XII, which is simply X + II. The number twenty seven 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. Input is guaranteed to be within the range from 1 to 3999.

Example 1:

Input: "III"
Output: 3

Example 2:

Input: "IV"
Output: 4

Example 3:

Input: "IX"
Output: 9

Example 4:

Input: "LVIII"
Output: 58
Explanation: L = 50, V= 5, III = 3.

Example 5:

Input: "MCMXCIV"
Output: 1994
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.

题解：

取当前char得到的数值若是比之前的小，就正常加上当前数值，若是比之前位大说明刚才加错了，需要先减掉再加上(cur-pre).

e.g. "XIV", at the beginning, res = 10. When i = 1, cur = 1, pre = 10, cur<pre, res = 11; when i = 2, cur = 5, pre = 1, cur>pre, res = 11-1+(5-1) = 14. 若是cur==pre，正常加cur, "XX" = 20.

Time Complexity: O(s.length()). Space: O(1).

AC Java:

 1 public class Solution {
 2     public int romanToInt(String s) {
 3         if(s == null || s.length() == 0){
 4             return 0;
 5         }
 6         
 7         int res = getValue(s.charAt(0));
 8         for(int i = 1; i<s.length(); i++){
 9             int pre = getValue(s.charAt(i-1));
10             int cur = getValue(s.charAt(i));
11             if(cur >pre){
12                 res = res-pre + (cur-pre);
13             }else{
14                 res += cur;
15             }
16         }
17         return res;
18     }
19     
20     private int getValue(char c){
21         int res = 0;
22         switch(c){
23             case 'I':
24                 res = 1;
25                 break;
26             case 'V':
27                 res = 5;
28                 break;
29             case 'X':
30                 res = 10;
31                 break;
32             case 'L':
33                 res = 50;
34                 break;
35             case 'C':
36                 res = 100;
37                 break;
38             case 'D':
39                 res = 500;
40                 break;
41             case 'M':
42                 res = 1000;
43                 break;
44             default:
45                 res = 0;
46         }
47         return res;
48     }
49 }

跟上Integer to Roman.