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: L = 50, V= 5, III = 3.
Example 5:
Input: "MCMXCIV" Output: 1994 Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.
羅馬數轉化成數字問題,我們需要對於羅馬數字很熟悉才能完成轉換。以下截自百度百科:
基本字元
|
I
|
V
|
X
|
L
|
C
|
D
|
M
|
相應的阿拉伯數字表示為
|
1
|
5
|
10
|
50
|
100
|
500
|
1000
|
class Solution { public: int romanToInt(string s) { int res = 0; unordered_map<char, int> m{{'I', 1}, {'V', 5}, {'X', 10}, {'L', 50}, {'C', 100}, {'D', 500}, {'M', 1000}}; for (int i = 0; i < s.size(); ++i) { int val = m[s[i]]; if (i == s.size() - 1 || m[s[i+1]] <= m[s[i]]) res += val; else res -= val; } return res; } };
我們也可以每次跟前面的數字比較,如果小於等於前面的數字,我們先加上當前的數字,比如 "VI",第二個字母 'I' 小於第一個字母 'V',所以要加1。如果大於的前面的數字,我們加上當前的數字減去二倍前面的數字,這樣可以把在上一個迴圈多加數減掉,比如 "IX",我們在 i=0 時,加上了第一個字母 'I' 的值,此時結果res為1。當 i=1 時,我們發現字母 'X' 大於前一個字母 'I',這說明前面的1是要減去的,而由於我們前一步不但沒減,還多加了個1,所以此時要減去2倍的1,就是減2,所以才能得到9,整個過程是 res = 1 + 10 - 2 = 9,參見程式碼如下:
解法二:
class Solution { public: int romanToInt(string s) { int res = 0; unordered_map<char, int> m{{'I', 1}, {'V', 5}, {'X', 10}, {'L', 50}, {'C', 100}, {'D', 500}, {'M', 1000}}; for (int i = 0; i < s.size(); ++i) { if (i == 0 || m[s[i]] <= m[s[i - 1]]) res += m[s[i]]; else res += m[s[i]] - 2 * m[s[i - 1]]; } return res; } };
類似題目:
參考資料:
https://leetcode.com/problems/roman-to-integer/
https://leetcode.com/problems/roman-to-integer/discuss/6547/Clean-O(n)-c%2B%2B-solution