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13. Roman to Integer

中文文档

Description

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, 2 is written as II in Roman numeral, just two ones added together. 12 is written as XII, which is simply X + II. The number 27 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.

 

Example 1:

Input: s = "III"
Output: 3
Explanation: III = 3.

Example 2:

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

Example 3:

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

 

Constraints:

  • 1 <= s.length <= 15
  • s contains only the characters ('I', 'V', 'X', 'L', 'C', 'D', 'M').
  • It is guaranteed that s is a valid roman numeral in the range [1, 3999].

Solutions

Solution 1: Hash Table + Simulation

First, we use a hash table $d$ to record the numerical value corresponding to each character. Then, we traverse the string $s$ from left to right. If the numerical value corresponding to the current character is less than the numerical value corresponding to the character on the right, we subtract the numerical value corresponding to the current character. Otherwise, we add the numerical value corresponding to the current character.

The time complexity is $O(n)$, and the space complexity is $O(m)$. Here, $n$ and $m$ are the length of the string $s$ and the size of the character set, respectively.

Python3

class Solution:
    def romanToInt(self, s: str) -> int:
        d = {'I': 1, 'V': 5, 'X': 10, 'L': 50, 'C': 100, 'D': 500, 'M': 1000}
        return sum((-1 if d[a] < d[b] else 1) * d[a] for a, b in pairwise(s)) + d[s[-1]]

Java

class Solution {
    public int romanToInt(String s) {
        String cs = "IVXLCDM";
        int[] vs = {1, 5, 10, 50, 100, 500, 1000};
        Map<Character, Integer> d = new HashMap<>();
        for (int i = 0; i < vs.length; ++i) {
            d.put(cs.charAt(i), vs[i]);
        }
        int n = s.length();
        int ans = d.get(s.charAt(n - 1));
        for (int i = 0; i < n - 1; ++i) {
            int sign = d.get(s.charAt(i)) < d.get(s.charAt(i + 1)) ? -1 : 1;
            ans += sign * d.get(s.charAt(i));
        }
        return ans;
    }
}

C++

class Solution {
public:
    int romanToInt(string s) {
        unordered_map<char, int> nums{
            {'I', 1},
            {'V', 5},
            {'X', 10},
            {'L', 50},
            {'C', 100},
            {'D', 500},
            {'M', 1000},
        };
        int ans = nums[s.back()];
        for (int i = 0; i < s.size() - 1; ++i) {
            int sign = nums[s[i]] < nums[s[i + 1]] ? -1 : 1;
            ans += sign * nums[s[i]];
        }
        return ans;
    }
};

Go

func romanToInt(s string) (ans int) {
	d := map[byte]int{'I': 1, 'V': 5, 'X': 10, 'L': 50, 'C': 100, 'D': 500, 'M': 1000}
	for i := 0; i < len(s)-1; i++ {
		if d[s[i]] < d[s[i+1]] {
			ans -= d[s[i]]
		} else {
			ans += d[s[i]]
		}
	}
	ans += d[s[len(s)-1]]
	return
}

TypeScript

function romanToInt(s: string): number {
    const d: Map<string, number> = new Map([
        ['I', 1],
        ['V', 5],
        ['X', 10],
        ['L', 50],
        ['C', 100],
        ['D', 500],
        ['M', 1000],
    ]);
    let ans: number = d.get(s[s.length - 1])!;
    for (let i = 0; i < s.length - 1; ++i) {
        const sign = d.get(s[i])! < d.get(s[i + 1])! ? -1 : 1;
        ans += sign * d.get(s[i])!;
    }
    return ans;
}

JavaScript

const romanToInt = function (s) {
    const d = {
        I: 1,
        V: 5,
        X: 10,
        L: 50,
        C: 100,
        D: 500,
        M: 1000,
    };
    let ans = d[s[s.length - 1]];
    for (let i = 0; i < s.length - 1; ++i) {
        const sign = d[s[i]] < d[s[i + 1]] ? -1 : 1;
        ans += sign * d[s[i]];
    }
    return ans;
};

C#

public class Solution {
    public int RomanToInt(string s) {
        Dictionary<char, int> d = new Dictionary<char, int>();
        d.Add('I', 1);
        d.Add('V', 5);
        d.Add('X', 10);
        d.Add('L', 50);
        d.Add('C', 100);
        d.Add('D', 500);
        d.Add('M', 1000);
        int ans = d[s[s.Length - 1]];
        for (int i = 0; i < s.Length - 1; ++i) {
            int sign = d[s[i]] < d[s[i + 1]] ? -1 : 1;
            ans += sign * d[s[i]];
        }
        return ans;
    }
}

PHP

class Solution {
    /**
     * @param String $s
     * @return Integer
     */
    function romanToInt($s) {
        $hashmap = [
            'I' => 1,
            'V' => 5,
            'X' => 10,
            'L' => 50,
            'C' => 100,
            'D' => 500,
            'M' => 1000,
        ];
        $rs = 0;
        for ($i = 0; $i < strlen($s); $i++) {
            $left = $hashmap[$s[$i]];
            $right = $hashmap[$s[$i + 1]];
            if ($left >= $right) {
                $rs += $left;
            } else {
                $rs -= $left;
            }
        }
        return $rs;
    }
}

Ruby

# @param {String} s
# @return {Integer}
def roman_to_int(s)
  hash = Hash[
      '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
  ]
  res = 0
  i = 0
  while i < s.length
    if i < s.length - 1 && !hash[s[i..i+1]].nil?
      res += hash[s[i..i+1]]
      i += 2
    else
      res += hash[s[i]]
      i += 1
    end
  end

  res
end