LeetCode--Roman to Integer
阿新 • • 發佈:2018-12-11
Roman to Integer
1. 題目
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
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
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.
2. 分析
這題實現起來不算難,只要注意一些規律,就是當前一個數字比後一個數字要小的時候,那麼這兩個數字必須是一個組合。然後map這個結構完美契合這個題目。
3. 原始碼
class Solution {
public:
int romanToInt(string s) {
map<char, int> m;
int sum = 0;
m['I'] = 1;
m['V'] = 5;
m['X'] = 10;
m['L'] = 50;
m['C'] = 100;
m['D'] = 500;
m['M'] = 1000;
for(int i = 0; i < s.size(); i++) {
int next = i+1;
if(next < s.size()) {
if(m[s[next]] > m[s[i]]) {
sum += (m[s[next]] - m[s[i]]);
i++;
} else {
sum += m[s[i]];
}
} else {
sum += m[s[i]];
}
}
return sum;
}
};