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.
Examples
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.
解题思路
和#12差不多，交换一下，从 int→罗马 到 罗马→int ，比较容易想到的方法就是一个一个对应咯
然后碰到需要转换的时候（当且仅当String中第i + 1个罗马字母代表的数字 > 第i个罗马字母代表的数字的时候才需要转换），就进行先-后+的操作就行啦
至于各个字母代表的数字可以通过map映射，不用每次都判断了，代码如下：
class Solution {
public int romanToInt(String s) {
int i = 0;
int temp = 0;
Map<Character, Integer> m = new HashMap<>();
m.put('I', 1);
m.put('V', 5);
m.put('X', 10);
m.put('L', 50);
m.put('C', 100);
m.put('D', 500);
m.put('M', 1000);
while(true){
if(i >= s.length() || i + 1 >= s.length())
break;
int a = m.get(s.charAt(i));
int b = m.get(s.charAt(i + 1));
if(a < b){
temp = temp + b - a;
i++;
}
else{
temp += a;
}
i++;
}
if(i == s.length() - 1)
temp += m.get(s.charAt(i));
return temp;
}
}
其他
- map里面的两个映射关系应该是object，不是数据类型（之前写了char，int是肯定过不了的）
- 往map中填充用的是put不是set QuQ
- 虽然是easy但是这次运行速度真的很快稍微有点小激动
- 继续刷！