LeetCode 13. Roman to Integer

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, 2 is written as II in Roman numeral, just two one's added together. 12 is written as XII, which is simply X + II. The number 27 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.

Example 1:

Input: s = "III"
Output: 3

Example 2:

Input: s = "IV"
Output: 4

Example 3:

Input: s = "IX"
Output: 9

Example 4:

Input: s = "LVIII"
Output: 58
Explanation: L = 50, V= 5, III = 3.

Example 5:

Input: s = "MCMXCIV"
Output: 1994
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.

Constraints:

1 <= s.length <= 15
s contains only the characters ('I', 'V', 'X', 'L', 'C', 'D', 'M').
It is guaranteed that s is a valid roman numeral in the range [1, 3999].

实现思路:
本题和12题正好对应,将罗马数字转为整数,方法就是用一个map容器保存相应的组成数,然后减法输出即可。

AC代码:

class Solution {
	public:
		int romanToInt(string s) {
			map<char,int> mp;
			mp['I']=1;
			mp['V']=5;
			mp['X']=10;
			mp['L']=50;
			mp['C']=100;
			mp['D']=500;
			mp['M']=1000;
			int size=s.length(),ans=0;
			for(int i=0; i<size; i++) {
				if(i+1<size) {
					if(mp[s[i+1]]>mp[s[i]]) {//当有下一个罗马字符存在并且小于前一位的字符时要做减法得值
						ans+=mp[s[i+1]]-mp[s[i]];
						i++;//这次要忽略两位  例如加了IV=4后前进2个位置
					} else ans+=mp[s[i]];//否则做加法
				} else ans+=mp[s[i]];//考虑到最后一个位置的情况
			}
			return ans;
		}
};
posted @ 2021-03-08 19:36  coderJ_ONE  阅读(39)  评论(0编辑  收藏  举报