地址
https://leetcode-cn.com/problems/integer-to-roman/
https://leetcode-cn.com/problems/roman-to-integer/submissions/
描述
思想
整数转罗马数字
罗马数字转整数:
定义映射,将单一字母映射到数字。
从前往后扫描,如果发现 s[i+1] 的数字比 s[i] 的数字大,那么累计 s[i+1]-s[i] 差值即可,并将 i 多向后移动一位;否则直接累计 s[i] 的值。
代码
class Solution {
public:
string intToRoman(int num) {
int values[] = {
1000,
900, 500, 400, 100,
90, 50, 40, 10,
9, 5, 4, 1
};
string reps[] = {
"M",
"CM", "D", "CD", "C",
"XC", "L", "XL", "X",
"IX", "V", "IV", "I"
};
string res;
for (int i = 0; i < 13; i ++ )
while(num >= values[i])
{
num -= values[i];
res += reps[i];
}
return res;
}
};
class Solution {
public:
int romanToInt(string s) {
unordered_map<char,int> map;
map['I']=1,map['V']=5,map['X']=10,map['L']=50,map['C']=100,map['D']=500,map['M']=1000;
int res=0;
for(int i=0;i<s.size();i++){
if(i+1<s.size()&&map[s[i]]<map[s[i+1]]){
res=res+map[s[i+1]]-map[s[i]];
i++;
}
else res+=map[s[i]];
}
return res;
}
};