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/*
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* @lc app=leetcode id=13 lang=cpp
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*
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* [13] Roman to Integer
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*
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* https://leetcode.com/problems/roman-to-integer/description/
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*
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* algorithms
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* Easy (53.27%)
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* Likes: 1444
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* Dislikes: 2872
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* Total Accepted: 488.4K
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* Total Submissions: 916.4K
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* Testcase Example: '"III"'
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*
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* Roman numerals are represented by seven different symbols: I, V, X, L, C, D
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* and M.
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*
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*
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* Symbol Value
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* I 1
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* V 5
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* X 10
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* L 50
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* C 100
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* D 500
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* M 1000
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*
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* For example, two is written as II in Roman numeral, just two one's added
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* together. Twelve is written as, XII, which is simply X + II. The number
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* twenty seven is written as XXVII, which is XX + V + II.
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*
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* Roman numerals are usually written largest to smallest from left to right.
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* However, the numeral for four is not IIII. Instead, the number four is
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* written as IV. Because the one is before the five we subtract it making
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* four. The same principle applies to the number nine, which is written as IX.
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* There are six instances where subtraction is used:
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*
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*
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* I can be placed before V (5) and X (10) to make 4 and 9.
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* X can be placed before L (50) and C (100) to make 40 and 90.
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* C can be placed before D (500) and M (1000) to make 400 and 900.
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*
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*
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* Given a roman numeral, convert it to an integer. Input is guaranteed to be
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* within the range from 1 to 3999.
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*
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* Example 1:
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*
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*
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* Input: "III"
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* Output: 3
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*
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* Example 2:
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*
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*
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* Input: "IV"
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* Output: 4
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*
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* Example 3:
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*
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*
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* Input: "IX"
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* Output: 9
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*
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* Example 4:
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*
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*
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* Input: "LVIII"
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* Output: 58
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* Explanation: L = 50, V= 5, III = 3.
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*
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*
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* Example 5:
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*
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*
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* Input: "MCMXCIV"
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* Output: 1994
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* Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.
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*
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*/
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#include <bits/stdc++.h>
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using namespace std;
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class Solution
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{
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public:
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int romanToInt(const string &s)
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{
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auto str = s.c_str();
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int res = 0, len = s.size();
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for (int i = 0; i < len; i++)
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switch (str[i])
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{
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case 'I':
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if (str[i + 1] == 'V')
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res += 4, i++;
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else if (str[i + 1] == 'X')
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res += 9, i++;
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else
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res += 1;
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break;
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case 'X':
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if (str[i + 1] == 'L')
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res += 40, i++;
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else if (str[i + 1] == 'C')
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res += 90, i++;
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else
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res += 10;
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break;
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case 'C':
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if (str[i + 1] == 'D')
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res += 400, i++;
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else if (str[i + 1] == 'M')
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res += 900, i++;
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else
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res += 100;
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break;
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case 'V':
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res += 5;
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break;
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case 'L':
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res += 50;
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break;
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case 'D':
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res += 500;
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break;
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case 'M':
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res += 1000;
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break;
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}
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return res;
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}
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};
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