Problem
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 beforeV
(5) andX
(10) to make 4 and 9.X
can be placed beforeL
(50) andC
(100) to make 40 and 90.C
can be placed beforeD
(500) andM
(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.
Solution
accumulate the value of each letter.
If the value of current letter is greater than the previous one, deduct twice of the previous value.
e.g. IX, 1 + 10 – 2 * 1 = 9 instead of 1 + 10 = 11
Time complexity: O(n)
Space complexity: O(1)
C++
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// Author: Huahua class Solution { public: int romanToInt(string s) { map<char, int> m{{'I', 1}, {'V', 5}, {'X', 10}, {'L', 50}, {'C', 100}, {'D', 500}, {'M', 1000}}; int ans = 0; for (int i = 0; i < s.length(); i++) { ans += m[s[i]]; if (i > 0 && m[s[i]] > m[s[i - 1]]) ans -= 2 * m[s[i - 1]]; } return ans; } }; |
Java
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// Author: Huahua class Solution { private int v(char R) { switch (R) { case 'I' : return 1; case 'V' : return 5; case 'X' : return 10; case 'L' : return 50; case 'C' : return 100; case 'D' : return 500; case 'M' : return 1000; default: break; } return 0; } public int romanToInt(String s) { int ans = 0; for (int i = 0; i < s.length(); i++) { ans += v(s.charAt(i)); if ( i > 0 && v(s.charAt(i)) > v(s.charAt(i - 1))) ans -= 2 * v(s.charAt(i - 1)); } return ans; } } |
Python3
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class Solution: def romanToInt(self, s): m = {'I' : 1, 'V': 5, 'X': 10, 'L' : 50, 'C' : 100, 'D': 500, 'M': 1000} ans = m[s[0]] for c, p in zip(s[1:], s[0:-1]): ans += m[c] if m[c] > m[p]: ans -= 2 * m[p] return ans |
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