[LeetCode]Roman to Integer

最新推荐文章于 2025-09-12 17:31:31 发布

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最新推荐文章于 2025-09-12 17:31:31 发布

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文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/Quincy/p/5315513.html

本文介绍了一种将罗马数字转换为整数的算法实现，包括解析罗马数字的规则及两种不同的转换方法，并通过测试案例验证了算法的有效性。

题目说明:

Given a roman numeral, convert it to an integer.

Input is guaranteed to be within the range from 1 to 3999.

程序代码:

#include <gtest/gtest.h>
using namespace std;

    //{'I','V','X','L','C','D','M','#','#'};
    //  1   5  10  50  100 500 1000

int getSingleRomanInt(char s)
{
    int nResult = 0;
    switch (s)
    {
    case 'I':
        nResult = 1;
        break;
    case 'V':
        nResult = 5;
        break;
    case 'X':
        nResult = 10;
        break;
    case 'L':
        nResult = 50;
        break;
    case 'C':
        nResult = 100;
        break;
    case 'D':
        nResult = 500;
        break;
    case 'M':
        nResult = 1000;
        break;
    }

    return nResult;
}

int romanToInt1(string s) 
{
    int nResult = 0;
    int nLength = s.length();
    if (nLength == 0)
    {
        return 0;
    }

    int nPre = getSingleRomanInt(s[0]);
    int nTemp = nPre;
    for (int i = 1; i < nLength; ++i)
    {
        int nCurrent = getSingleRomanInt(s[i]);
        if (nPre == nCurrent)
        {
            nTemp += nCurrent;
        }
        else if(nPre < nCurrent)
        {
            nTemp = nCurrent - nTemp;
        }
        else
        {
            nResult += nTemp;
            nTemp = nCurrent;
        }

        nPre = nCurrent;
    }

    nResult += nTemp;
    return nResult;
}

// 虽然罗马数字的书写规则较为复杂，但根据罗马数字“左加右减”的规律，
// 可以构造出更简单的罗马数字转换阿拉伯数字的方法：即从右向左（从低位向高位）考察罗马数字，
// 遇到比上一个数字大的数字就加上，遇到比上一个数字小的数字就减去。
int romanToInt(string s) 
{
    int nResult = 0;
    int nLength = s.length();
    if (nLength == 0)
    {
        return 0;
    }

    int nPre = 0;
    for (int i = nLength-1; i >=0; i--)
    {
        int nCurrent = getSingleRomanInt(s[i]);
        if (nCurrent < nPre)
        {
            nResult -= nCurrent;
            nPre = nCurrent;
        }
        else
        {
            nResult += nCurrent;
            nPre = nCurrent;
        }
    }

    return nResult;
}

TEST(Pratices, tRomanToInt)
{
    // DLLXXXVII => 687
    ASSERT_EQ(romanToInt("DCLXXXVII"),687);

    // CCI => 201
    ASSERT_EQ(romanToInt("CCI"),201);

    // MMMCMXCIX => 3999
    ASSERT_EQ(romanToInt("MMMCMXCIX"),3999);

    // CDLVI => 456
    ASSERT_EQ(romanToInt("CDLVI"),456);

    // XCIX => 99
    ASSERT_EQ(romanToInt("XCIX"),99);

    ASSERT_EQ(romanToInt("IV"),4);
}