【LeetCode & 剑指offer刷题】发散思维题8：Roman to Integer

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Roman to Integer

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: C = 100, L = 50, XXX = 30 and III = 3.

Example 5:

Input: "MCMXCIV"

Output: 1994

Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.

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C++

 

//问题：罗马数字转成整数

/*

Roman numerals are usually written largest to smallest from left to right.

一般从大到小写，如果从小到大写为减法，如IV（4） IX（9）

*/

#include <unordered_map>

#include <string>

class Solution

{

public:

    int romanToInt(string s)

    {

        unordered_map<char, int> map =  { { 'I' , 1 },

                                   { 'V' , 5 },

                                   { 'X' , 10 },

                                   { 'L' , 50 },

                                   { 'C' , 100 },

                                   { 'D' , 500 },

                                   { 'M' , 1000 } };

        int sum = map[s.back()]; //字符串末尾字符

        for(int i = s.size()-2; i>=0; i--) //从后往前扫描

        {

            if(map[s[i]] < map[s[i+1]]) sum -= map[s[i]]; //当前字符比后面字符对应数值小时，用减法

            else sum += map[s[i]];

        }

        return sum;

    }

};