【leetcode】13. Roman to Integer

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;
}

代码2

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;
}