【leetcode】13. Roman to Integer
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一、题目描述
Given a roman numeral, convert it to an integer.
Input is guaranteed to be within the range from 1 to 3999.
题目解读:将罗马数字转化为对应的整数
I II III IV V VI VII VIII IX X
1 2 3 4 5 6 7 8 9 10
I 1
V 5
X 10
L 50
C 100
D 500
M 1000
思路:设置三个标志,比如IV是4 , IX是9,XL是40,CD是400,CM是900
所以当出现I,X,C时要标记一下,在判断下一个字符的时候要将这三个是否出现在前面考虑进来。
c++代码(36ms)
class Solution {public: int romanToInt(string s) { int result = 0; int len = s.length(); int flag1 = 0; //I的标记 int flag2 = 0; //X的标记 int flag3 = 0; //C的标记 for(int i=0; i<len; i++){ if(s[i] == 'I'){ result += 1; flag1 = 1; //标记前一个是I } else if(s[i] == 'V'){ //5 if(flag1){ result += 3; flag1 = 0; } else{ result += 5; } } else if(s[i] == 'X'){ //10 if(flag1){ result += 8; flag1 = 0; } else{ flag2 = 1; result += 10; } } else if(s[i] == 'L'){ if(flag2){ result += 30; flag2 = 0; } else result += 50; } else if(s[i] == 'C'){ if(flag2){ result += 80; flag2 = 0; } else{ flag3 = 1; result += 100; } } else if(s[i] == 'D'){ if(flag3){ result += 300; flag3 = 0; } else result += 500; } else if(s[i] == 'M'){ if(flag3){ result += 800; flag3 = 0; } else result += 1000; } } return result; }};
注意:
当出现错误
invalid conversion from ‘const char*’ to ‘const char&’ [-fpermissive]
时,将字符的" "改成' '(双引号改成单引号)即可。
其他代码,使用map存I,V,X,L,C,D,M的值
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 romanToInt(string s) { if (s.empty()) return 0; int roman[24] = {}; roman['I' - 'A'] = 1; roman['V' - 'A'] = 5; roman['X' - 'A'] = 10; roman['L' - 'A'] = 50; roman['C' - 'A'] = 100; roman['D' - 'A'] = 500; roman['M' - 'A'] = 1000; auto sum = 0; auto right = roman[s.front() - 'A']; for (int i = 1; i < s.size(); ++i) { auto curr = right; right = roman[s[i] - 'A']; if (right > curr) sum -= curr; else sum += curr; } return sum + right;}
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