LeetCode: 13. Roman to Integer

051106

题目

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, 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 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.

题意

有7个这样的罗马字母,分别代表7个数字。一般来说,写法是从大到小写为从左到右的,然后依次相加。但是为了更简洁的表示一些大值,当写法为从小到大时,则表示相减。

我的解题思路

首先建立一个字符到数值的map,然后判断它们的位置和大小,如果大的在前就直接加大的,如果小的在前,则加(大值-小值),注意对最后一个字符的处理。

class Solution {
public:
    int romanToInt(string s) {
        unordered_map<char, int> map(7);
        map['I']=1;
        map['V']=5;
        map['X']=10;
        map['L']=50;
        map['C']=100;
        map['D']=500;
        map['M']=1000;
        
        if(s.length()==1) return map[s[0]];
        
        int res=0;
        for(int i=0; i<s.length()-1; i++){
            if(map[s[i]]>=map[s[i+1]]) {
                res+=map[s[i]];
            }
            else{
                res+=(map[s[i+1]]-map[s[i]]);
                i++; 
            }
            if(i==s.length()-2) res+=map[s[i+1]];
        }
        return res;     
    }
};

看了下高赞解答,发现思路跟我也差不多。

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转载自blog.csdn.net/liveway6/article/details/90108698