题目描述(Easy)
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.
题目链接
https://leetcode.com/problems/roman-to-integer/description/
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.
算法分析
从前往后扫描，如果当前数字大于上一个数字，则加上当前数值减去上一个数值的值；否则，将当前值加入结果。
提交代码：
class Solution {
public:
inline int map(const char c) {
switch (c) {
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: return 0;
}
}
int romanToInt(string s) {
int result = 0;
result += map(s[0]);
for (int i = 1; i < s.size(); ++i)
{
if(map(s[i]) > map(s[i - 1]))
result += map(s[i]) - 2 * map(s[i - 1]);
else
result += map(s[i]);
}
return result;
}
};测试代码：
// ====================测试代码====================
void Test(const char* testName, string str, int expected)
{
if (testName != nullptr)
printf("%s begins: \n", testName);
Solution s;
int result = s.romanToInt(str);
if(result == expected)
printf("passed\n");
else
printf("failed\n");
}
int main(int argc, char* argv[])
{
Test("Test1", string("III"), 3);
Test("Test2", string("IV"), 4);
Test("Test3", string("IX"), 9);
Test("Test4", string("LVIII"), 58);
Test("Test5", string("MCMXCIV"), 1994);
return 0;
}