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Given a roman numeral, convert it to an integer.
Input is guaranteed to be within the range from 1 to 3999.
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按照intger to roman 的思路,写下面这道题,比较绕, 而且写的不够简洁,而且修改了好几次
class Solution { public: int romanToInt(string s) { string c[4][10]={{"0","I","II","III","IV","V","VI","VII","VIII","IX"}, {"0","X","XX","XXX","XL","L","LX","LXX","LXXX","XC"}, {"0","C","CC","CCC","CD","D","DC","DCC","DCCC","CM"}, {"0","M","MM","MMM"}}; int len = s.length(); if (0 == len) { return 0; } int begin = 0; int i; int res = 0; while (begin < len) { if (s[begin] == ‘M‘) { for (i=3; i>0; i--) { if (s.find(c[3][i]) != string::npos) { res += i * 1000; begin = begin + c[3][i].length(); break; } } } else if (s[begin] == ‘C‘ || s[begin] == ‘D‘) { for (i=9; i>0; i--) { if (s.find(c[2][i]) != string::npos) { //当找到D的时候,并不能够确定是找到CD 还是D //做一下判断 if (begin >=0 && s[begin] == ‘C‘ && s[begin + 1] == ‘D‘) { res += 4 * 100; begin = begin + c[2][4].length(); break; } else { res += i * 100; begin = begin + c[2][i].length(); break; } } } } else if (s[begin] == ‘X‘ || s[begin] == ‘L‘) { for (i=9; i>0; i--) { if (s.find(c[1][i]) != string::npos) { if (begin >=0 && s[begin] == ‘X‘ && s[begin + 1] == ‘L‘) { res += 4 * 10; begin = begin + c[1][4].length(); break; } else { res += i * 10; begin = begin + c[1][i].length(); break; } } } } else if (s[begin] == ‘I‘ || s[begin] == ‘V‘) { for (i=9; i>0; i--) { if (s.find(c[0][i]) != string::npos) { if (begin >=0 && s[begin] == ‘I‘ && s[begin + 1] == ‘V‘) { res += 4 ; begin = begin + c[0][4].length(); break; } else { res += i; begin = begin + c[0][i].length(); break; } } } } else { return res; } } return res; } };
如果按照下面这种思路写就很简单了
class Solution { public: int romanToInt(string s) { map<char,int> roman; roman[‘I‘]=1; roman[‘V‘]=5; roman[‘X‘]=10; roman[‘L‘]=50; roman[‘C‘]=100; roman[‘D‘]=500; roman[‘M‘]=1000; int len = s.length(); if (0 == len) { return 0; } int res = 0; int index; for (index=0; index<len; index++) { if (index>0 && roman[s[index]] > roman[s[index -1]]) { res = res + (roman[s[index]] -2 * roman[s[index -1]]); } else { res = res + roman[s[index]]; } } return res; } };
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原文地址:http://www.cnblogs.com/SpeakSoftlyLove/p/5097151.html