From 780c10ecd0cae0800c93b20ffe778fd072359dfb Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E5=94=90=E6=A0=91=E6=A3=AE?= <14021051@buaa.edu.cn>
Date: Thu, 27 Sep 2018 17:52:06 +0800
Subject: [PATCH] Create 13. Roman to Integer.md

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+# [13. Roman to Integer](https://leetcode.com/problems/roman-to-integer/description/)
+# 思路
+罗马数字转阿拉伯数字。七个大写字母对应七个数，对应关系可以用一个数组（或者map）来实现。  
+用res代表结果，将代表罗马数字的字符串s从前往后遍历，如果当前字母对应的数大于等于下一个字母对应的数，则res加上对应的数，否则减。   
+注意：根据题意是没有可能发生s[i] < s[i + 1] < s[i + 2]的。
+# C++
+```
+class Solution {
+public:
+    int romanToInt(string s) {
+        vector<int>mp(26);
+        mp['I' - 'A'] = 1;
+        mp['V' - 'A'] = 5;
+        mp['X' - 'A'] = 10;
+        mp['L' - 'A'] = 50;
+        mp['C' - 'A'] = 100;
+        mp['D' - 'A'] = 500;
+        mp['M' - 'A'] = 1000;
+        int res = mp[s[s.size() - 1] - 'A'];
+        for(int i = 0; i < s.size() - 1; i++){
+            if(mp[s[i] - 'A'] < mp[s[i + 1] - 'A']) res -= mp[s[i] - 'A'];
+            else res += mp[s[i] - 'A'];
+        }
+        return res;
+    }
+};
+```