【leetcode】Roman to Integer

來源:互聯網
上載者:User

Question :

Given a roman numeral, convert it to an integer.

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

Anwser 1 :       

class Solution {public:    int toNum(char c)      {          switch(c)          {              case 'I':                  return 1;              case 'V':                  return 5;              case 'X':                  return 10;              case 'L':                  return 50;              case 'C':                  return 100;              case 'D':                  return 500;              case 'M':                  return 1000;              default:                  return 0;          }      }         int romanToInt(string s) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        int ret = 0;    for(int i = 0; i < s.size(); i++){if(i + 1 < s.size()){if(toNum(s[i]) < toNum(s[i+1]))     // high num < low num, then minus{ret -= toNum(s[i]);}else{ret += toNum(s[i]);}}else{ret += toNum(s[i]);}}return ret;    }};

Anwser 2 :        

class Solution {public:    int toNum(char c)      {          switch(c)          {              case 'I':                  return 1;              case 'V':                  return 5;              case 'X':                  return 10;              case 'L':                  return 50;              case 'C':                  return 100;              case 'D':                  return 500;              case 'M':                  return 1000;              default:                  return 0;          }      }         int romanToInt(string s) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        int ret = 0;                int len = s.size();        int pre = 1001;        int cur;        for (int i = 0; i < len; ++i)        {            cur = toNum(s[i]);            ret += cur;            if (cur > pre)            {                ret -= 2*pre;   // pre already add one time, now minus tow times            }            pre = cur;        }        return ret;    }};

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