13. Roman to Integer*

​​https://leetcode.com/problems/roman-to-integer/​​

题目描述

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.

题目描述

本题是 12. Integer to Roman**​ 的逆变换. 仍旧使用哈希表, 当访问到 ​​s[i]​​​ 时, 判断 ​​s[i, i+1]​​​ 这个子字符串是否在哈希表中 (比如 “IV”, “CM”). 如果不存在, 说明当前 ​​s[i]​​ 只需要用一个字符来表示一个数, 而不需要用两个字符来表示一个数.

class Solution {
public:
    int romanToInt(string s) {
        unordered_map<string, int> record{{"I", 1}, {"V", 5}, {"X", 10},
                                          {"L", 50}, {"C", 100}, {"D", 500},
                                          {"M", 1000}, {"IV", 4}, {"IX", 9},
                                          {"XL", 40}, {"XC", 90}, {"CD", 400},
                                          {"CM", 900}};
        int res = 0;
        for (int i = 0; i < s.size(); ++ i) {
            if (i + 1 < s.size() && record.count(s.substr(i, 2))) {
                res += record[s.substr(i, 2)];
                ++ i;
            } else {
                res += record[string(1, s[i])];
            }
        }
        return res;
    }
};

C++ 实现 2

两年前的代码.

class Solution {
private:
    unordered_map<string, int> symbols = {
        {"I", 1},
        {"IV", 4},
        {"V", 5},
        {"IX", 9},
        {"X", 10},
        {"XL", 40},
        {"L", 50},
        {"XC", 90},
        {"C", 100},
        {"CD", 400},
        {"D", 500},
        {"CM", 900},
        {"M", 1000},
    };
public:
    int romanToInt(string s) {
        int res = 0;
        for (int i = 0; i < s.size(); ) {
            // s.substr(pos, len), 如果 len 超过了剩余的字符串的长度,
            // 那么就返回剩余的字符串, 所以这里不需要对 i 的范围做检查
            if (!symbols.count(s.substr(i, 2))) {
                res += symbols[s.substr(i, 1)];
                ++ i;
            }
            else {
                res += symbols[s.substr(i, 2)];
                i += 2;
            }
        }
        return res;
    }
};