[LeetCode] Roman to Integer

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Roman to Integer

Given a roman numeral, convert it to an integer.

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

解题思路：

罗马数字转化成数字。一种方法是映射法，每一位（阿拉伯数字）的字符到阿拉伯数字的映射。如下所示：

class Solution {public:    int romanToInt(string s) {        int result=0;                map<string, int> m({            {"M", 1000}, {"MM",2000}, {"MMM", 3000},            {"C", 100}, {"CC", 200}, {"CCC", 300}, {"CD", 400}, {"D", 500}, {"DC", 600}, {"DCC", 700}, {"DCCC", 800}, {"CM", 900},            {"X", 10}, {"XX", 20}, {"XXX", 30}, {"XL", 40}, {"L", 50}, {"LX", 60}, {"LXX", 70}, {"LXXX", 80}, {"XC", 90},            {"I", 1}, {"II", 2}, {"III", 3}, {"IV", 4}, {"V", 5}, {"VI", 6}, {"VII", 7}, {"VIII", 8}, {"IX", 9}        });                int len=s.length();        int end=0, start=0;        string str;                //找到千位        while(s[end]=='M' && end<len){            end++;        }        if(end!=start){            str=s.substr(start, end-start);            result += m[str];            start=end;        }                //找到百位        while((s[end]=='C'||s[end]=='M'||s[end]=='D') && end<len){            end++;        }        if(end!=start){            str=s.substr(start, end-start);            result += m[str];            start=end;        }                //找到十位        while((s[end]=='X'||s[end]=='L'||s[end]=='C') && end<len){            end++;        }        if(end!=start){            str=s.substr(start, end-start);            result += m[str];            start=end;        }                //找到个位        while((s[end]=='I'||s[end]=='V'||s[end]=='X') && end<len){            end++;        }        if(end!=start){            str=s.substr(start, end-start);            result += m[str];            start=end;        }                return result;    }};

上述比较麻烦，感觉自己是老黄牛。另外一种方法，从左往右观察罗马字符，若前一个字符转化为阿拉伯数字比后一个字符转化为阿拉伯数字大，则直接将前一个字符加到结果中，否则将后一个字符减去前一个字符的差加入到结果中。这种方法非常clever。

class Solution {public:    int romanToInt(string s) {        int len=s.length();        if(len==0){            return 0;        }                int result=0;                int lastNum = 0;                for(int i=0; i<len; i++){            int thisNum = c2n(s[i]);            if(lastNum==0){                lastNum=thisNum;            }else if(lastNum<thisNum){                result += thisNum - lastNum;                lastNum=0;            }else{                result += lastNum;                lastNum = thisNum;            }        }                result += lastNum;                        return result;    }        int c2n(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;          }    }};

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