13 Roman to Integer 羅馬數字轉整數
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, 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:
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.
題目描述:
羅馬數字包含以下七種字符: I， V， X， L，C，D 和 M。
字符 數值
I 1
V 5
X 10
L 50
C 100
D 500
M 1000
例如， 羅馬數字 2 寫做 II ，即為兩個并列的 1。12 寫做 XII ，即為 X + II 。 27 寫做 XXVII, 即為 XX + V + II 。
通常情況下，羅馬數字中小的數字在大的數字的右邊。但也存在特例，例如 4 不寫做 IIII，而是 IV。數字 1 在數字 5 的左邊，所表示的數等于大數 5 減小數 1 得到的數值 4 。同樣地，數字 9 表示為 IX。這個特殊的規則只適用于以下六種情況：
I 可以放在 V (5) 和 X (10) 的左邊，來表示 4 和 9。
X 可以放在 L (50) 和 C (100) 的左邊，來表示 40 和 90。
C 可以放在 D (500) 和 M (1000) 的左邊，來表示 400 和 900。
給定一個羅馬數字，將其轉換成整數。輸入確保在 1 到 3999 的范圍內。
示例:
示例 1:
輸入: "III"
輸出: 3
示例 2:
輸入: "IV"
輸出: 4
示例 3:
輸入: "IX"
輸出: 9
示例 4:
輸入: "LVIII"
輸出: 58
解釋: L = 50, V= 5, III = 3.
示例 5:
輸入: "MCMXCIV"
輸出: 1994
解釋: M = 1000, CM = 900, XC = 90, IV = 4.
思路:
- 建立一個map, 映射羅馬數字和整數即可, 將所在位置與其后一個位置比較選擇加或者減
- 由于IV可以看做 1 + 5 - 2, 可以先查找減去相應值再相加
時間復雜度O(n), 空間復雜度O(1)
代碼:
C++:
class Solution
{
public:
int romanToInt(string s)
{
int result = 0;
if (s.find("IV") != -1 or s.find("IX") != -1) result -= 2;
if (s.find("XL") != -1 or s.find("XC") != -1) result -= 20;
if (s.find("CD") != -1 or s.find("CM") != -1) result -= 200;
for (int i = 0; i < s.length(); i++)
{
switch (s[i])
{
case 'I':
result += 1;
break;
case 'V':
result += 5;
break;
case 'X':
result += 10;
break;
case 'L':
result += 50;
break;
case 'C':
result += 100;
break;
case 'D':
result += 500;
break;
case 'M':
result += 1000;
break;
}
}
return result;
}
};
Java:
class Solution {
public int romanToInt(String s) {
int result = 0;
if (s.contains("IV") || s.contains("IX")) {
result -= 2;
}
if (s.contains("XL") || s.contains("XC")) {
result -= 20;
}
if (s.contains("CD") || s.contains("CM")) {
result -= 200;
}
for (int i = 0; i < s.length(); i++) {
switch (s.charAt(i)) {
case 'I':
result += 1;
break;
case 'V':
result += 5;
break;
case 'X':
result += 10;
break;
case 'L':
result += 50;
break;
case 'C':
result += 100;
break;
case 'D':
result += 500;
break;
case 'M':
result += 1000;
break;
}
}
return result;
}
}
Python:
class Solution:
def romanToInt(self, s: str) -> int:
roman_dic = {'I':1, 'V':5, 'X':10, 'L':50, 'C':100, 'D':500, 'M':1000}
result = 0
length = len(s)
for i in range(length):
if i < length - 1 and roman_dic[s[i]] < roman_dic[s[i + 1]]:
result -= roman_dic[s[i]]
else:
result += roman_dic[s[i]]
return result
