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  • LeetCode

    13. Roman to Integer

    Problem's Link

     ----------------------------------------------------------------------------

    Mean: 

    给你一个字符串,代表罗马数字,将其转换为int型数字.

    analyse:

    Time complexity: O(N)

     

    view code

    /**
    * -----------------------------------------------------------------
    * Copyright (c) 2016 crazyacking.All rights reserved.
    * -----------------------------------------------------------------
    *       Author: crazyacking
    *       Date  : 2016-02-16-12.06
    */
    #include <queue>
    #include <cstdio>
    #include <set>
    #include <string>
    #include <stack>
    #include <cmath>
    #include <climits>
    #include <map>
    #include <cstdlib>
    #include <iostream>
    #include <vector>
    #include <algorithm>
    #include <cstring>
    #include <bits/stdc++.h>
    using namespace std;
    typedef long long(LL);
    typedef unsigned long long(ULL);
    const double eps(1e-8);

    class Solution
    {
    public:
       int romanToInt(string s)
       {
           unordered_map<char, int> T =
           {
               { 'I' , 1   },
               { 'V' , 5   },
               { 'X' , 10  },
               { 'L' , 50  },
               { 'C' , 100 },
               { 'D' , 500 },
               { 'M' , 1000}
           };

           int sum = T[s.back()];
           for (int i = s.length() - 2; i >= 0; --i)
           {
               if (T[s[i]] < T[s[i + 1]])
                   sum -= T[s[i]];
               else
                   sum += T[s[i]];
           }
           return sum;
       }
    };


    int main()
    {
       Solution solution;
       string s;
       while(cin>>s)
       {
           cout<<solution.romanToInt(s)<<endl;
       }
       return 0;
    }
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  • 原文地址:https://www.cnblogs.com/crazyacking/p/5035757.html
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