# LeetCode刷题记录——day5

class Solution {
public:
int romanToInt(string s) {
unordered_map<string, int> m = {{"I", 1}, {"IV", 3}, {"IX", 8}, {"V", 5}, {"X", 10}, {"XL", 30}, {"XC", 80}, {"L", 50}, {"C", 100}, {"CD", 300}, {"CM", 800}, {"D", 500}, {"M", 1000}};
int r = m[s.substr(0, 1)];
for(int i=1; i<s.size(); ++i){
string two = s.substr(i-1, 2);
string one = s.substr(i, 1);
r += m[two] ? m[two] : m[one];
}
return r;
}
};

class Solution {
public:
string intToRoman(int num) {
int len=0,i=0;
string s;
len = num/1000;
num=num%1000;
for(;i<len;i++){
s+='M';
}

len=len+2*(num/900);
num=num%900;
for(;i<len;i=i+2){
s+='C';
s+='M';
}

len=len+num/500;
num=num%500;
for(;i<len;i++){
s+='D';
}

len=len+2*(num/400);
num=num%400;
for(;i<len;i=i+2){
s+='C';
s+='D';
}

len=len+num/100;
num=num%100;
for(;i<len;i++){
s+='C';
}

len=len+2*(num/90);
num=num%90;
for(;i<len;i=i+2){
s+='X';
s+='C';
}

len=len+num/50;
num=num%50;
for(;i<len;i++){
s+='L';
}

len=len+2*(num/40);
num=num%40;
for(;i<len;i=i+2){
s+='X';
s+='L';
}

len=len+num/10;
num=num%10;
for(;i<len;i++){
s+='X';
}

len=len+2*(num/9);
num=num%9;
for(;i<len;i=i+2){
s+='I';
s+='X';
}

len=len+(num/5);
num=num%5;
for(;i<len;i++){
s+='V';
}

len=len+2*(num/4);
num=num%4;
for(;i<len;i=i+2){
s+='I';
s+='V';
}

len=len+num;
for(;i<len;i++){
s+='I';
}

return s;
}
};


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;
int
for (int i = 0; i < 13; i ++ )  //这里不使用图里的count了，一遍一遍来就行了
while(num >= values[i])
{
num -= values[i];
res += reps[i];
}
return res;
}
};

posted @ 2024-03-23 20:41  想成为编程高手的阿曼  阅读(2)  评论(0编辑  收藏  举报