LeetCode [Roman to Integer]

最新推荐文章于 2019-02-12 10:37:52 发布

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本文详细介绍了一种将罗马数字转换为整数的算法，特别关注了4和9的特殊表示，提出了一种通过比较相邻字符值并进行相应加减运算的解决方案。文章提供了C和C++的实现代码，适合对算法和编程感兴趣的读者。

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.

**问题解析：**大概的题意就是给出罗马数字让翻译成阿拉伯数字，重点是4， 9的表示不符合常规，我们需要特殊“照顾"
**思路：**我们可以设一个ans来记录罗马数对应的阿拉伯数值（初值为0），从字符最后一个遍历，如果第i个字符比它前一个字符大，则ans加上第i个字符对应的数值减去它前一个字符对应的数值，否则就加上本身。例如MCMXCIV，V比I大，则ans+=5-1（4）。C比X大，则ans+=100-10（94）.以此类推。
C语言代码如下：

#include<cstdio>
#include<cmath>
#include<map>
#include<algorithm>
#include<iostream>
#include<cstring>
using namespace std;
int main()
{
	char str[20];
	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;
	cin >> str;
	int n=strlen(str);
	int ans=0;
	for(int i = n-1;i>=0;i--)	
	{
		if(i==0)
		{
			ans+=mp[str[i]];
			break;
		}
		else
		{
				if(mp[str[i]]>mp[str[i-1]])
			{
				ans+=mp[str[i]]-mp[str[i-1]];
				i--;
			}
			else
			{
				ans+=mp[str[i]];
			}
		}
		
		
	}
	cout<<ans;


 return 0;
}

LeetCode里由于C不能用map函数，因此用C+++提交，代码如下：

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 ans=0;
        for(int i = s.size();i>=0;i--)	
        {
            if(i==0)
            {
                ans+=mp[s[i]];
                break;
            }
            else
            {
                    if(mp[s[i]]>mp[s[i-1]])
                {
                    ans+=mp[s[i]]-mp[s[i-1]];
                    i--;
                }
                else
                {
                    ans+=mp[s[i]];
                }
            }


        }
        return ans;
        
    }
};

大家可以看一下C++中string的用法：标准C++中的string类的用法总结
不熟悉map函数的童鞋可以去搜一下，这里不再多讲，附上一个map函数的简单应用map函数统计字符个数