LeetCodet题解--13. Roman to Integer

链接

LeetCode题目：https://leetcode.com/problems/roman-to-integer/

GitHub代码：https://github.com/gatieme/LeetCode/tree/master/013-RomanToInteger

CSDN题解：http://blog.csdn.net/gatieme/article/details/51052713

题意

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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与12题正好相反，这道题把一个不大于4000的罗马数字转换为十进制整数

所有的可能性–按位翻译

因为数字不大于4000，因此我们可以分析罗马数字个位、十位、百位、千位的规则，然后计算所有的可能性然后分析

string one[11] = {"", "I","II","III","IV","V","VI","VII","VIII","IX"};          ///  个位
string ten[11] = {"", "X", "XX","XXX","XL","L","LX","LXX","LXXX","XC"};         ///  十位
string hundred[11] = {"", "C","CC","CCC","CD","D","DC","DCC","DCCC","CM"};      ///  百位
string thousand[4] = {"", "M", "MM","MMM"};

因此有了这样的代码

#include <iostream>

using namespace std;

/**

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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string one[11] = {"", "I","II","III","IV","V","VI","VII","VIII","IX"};          ///  个位
string ten[11] = {"", "X", "XX","XXX","XL","L","LX","LXX","LXXX","XC"};         ///  十位
string hundred[11] = {"", "C","CC","CCC","CD","D","DC","DCC","DCCC","CM"};      ///  百位
string thousand[4] = {"", "M", "MM","MMM"};                                  ///  千位

*/

class Solution {
public:
int romanToInt(string s)
{
int result = 0;
const char* p = s.c_str( );
int count = 0;

//千位
while(*p!='\0'&&*p=='M')
{
p++;        //千位循环结束后p指向百位。
count++;
}
result += 1000 * count;
count=0;

//百位
if(*p == 'C')
{
while(*p!='\0'&&*p=='C')
{
p++;
count++;
}
if(count==1&&*p=='M')
{
p++;
count=9;
}
else if(count==1&&*p=='D')
{
p++;
count=4;
}
}else if(*p=='D')
{
p++;
count=5;
while(*p!='\0'&&*p=='C')
{

p++;
count++;
}

}
else
{
count=0;
}
result+=count*100;
count=0;

//十位
if(*p=='X')
{
while(*p!='\0'&&*p=='X')
{
p++;
count++;
}
if(count==1&&*p=='C')
{
p++;
count=9;
}
else if(count==1&&*p=='L')
{
p++;
count=4;
}
}else if(*p=='L') {
p++;
count=5;
while(*p!='\0'&&*p=='X')
{
p++;
count++;
}

}else{
count=0;
}
result+=count*10;
count=0;

//个位
if(*p=='I')
{
while(*p!='\0'&&*p=='I')
{
p++;
count++;
}
if(count==1&&*p=='X')
{
p++;
count=9;
}
else if(count==1&&*p=='V')
{
p++;
count=4;
}
}
else if(*p=='V')
{
p++;
count=5;
while(*p!='\0'&&*p=='I')
{
p++;
count++;
}

}
else
{
count=0;
}
result+=count;

return result;
}
};

分析–寻找罗马数字的规则

上面的解法太繁琐太深入细节了，观察到罗马数字和Integer的转换的小规律是：

IV = 5 - 1 = (-1) + 5 = 4

VI = 5 + 1 = 5 + 1 = 6

I在V前面，由于I比V小，所以I被解释为负数。

V在I后面，由于V比I大，所以V给解释为整数。

继续看几个例子。

VII = 5 + 1 + 1 = 7

IX = (-1) + 10 = 9

所以，

扫描字符串一遍，从第一个字符开始，到最后一个字符为止，一次比较当前字符a和当前字符的下一个字符b。

如果a< b ，解释为 b - a,否则如果a >= b， 解释为a + b。

实际上，由于每次总是比较2个相邻的字符，所以是下标是从0开始，到

inputstring.length()-2

结束。

class Solution {
public:
int romanToInt(std::string s)
{
int sum = 0;
int start = 0;

char char_arr[] = {'I', 'V', 'X', 'L', 'C', 'D', 'M'};
int int_arr[] = {1, 5, 10, 50, 100, 500, 1000};

std::map<char, int> roman_map;
int map_len = sizeof(char_arr) / sizeof(char);

for (int i = 0; i< map_len; ++i)
{
roman_map.insert(std::pair<char, int> (char_arr[i], int_arr[i]));
}
for (int i = 0; i < s.length() - 1; ++i)
{
if (roman_map[s[i]]>=roman_map[s[i + 1]])
{
sum += roman_map[s[i]];
}
else
{

sum -= roman_map[s[i]];
}
}
sum += roman_map[s[s.length()-1]];
return sum;
}
};