roman to integer

class Solution {
public:
int romanToInt(string s) {
if(s.size() == 0) return 0;
int re = 0;
//index 表示当前应该处理的元素。
int index = 0;
int len = s.size();
//处理跟M有关的数
int thousand = s.find_first_of('M');
if (-1 != thousand ){//find
if( thousand == 0){//it's "M****"
//处理连续M,连续则累加。
while(index < len && s[index] == 'M'){
re += 1000;
++index;
}
//出现不连续的M，则是900.
thousand = s.find_last_of('M');
if (thousand > index){
re += 900;
index = thousand + 1;
}
if(index == len) return re;
}
else{//M没在第一位，则表示的数是介于[900,1000)
re += 900;
index = thousand + 1;
}
}
//处理跟D有关的数，[400,900)
int five_hun = s.find_first_of('D');
if ( -1 != five_hun){
if(five_hun == index){// >= 500
re += 500;
index++;
}
else{// it must be "CD"
re += 400;
index = five_hun + 1;
}
}
//处理跟C有关的数[90, )
int hun = s.find_first_of('C');
if (hun >= index){
if (hun == index){
//C前面没有X，[100, )
while (index < len && s[index] == 'C'){
re += 100;
++index;
}
hun = s.find_last_of('C');
if (hun > index){
re += 90;
index = hun + 1;
}
if(index == len) return re;
}
else{//it must be "XC"
re += 90;
index = hun + 1;
}
}

//handle 'L'
int five_ten = s.find_first_of('L');
if (five_ten >= index){
if (five_ten == index){//"V***"
re += 50;
index++;
if(index == len) return re;
}
else{//"XV"
re += 40;
index = five_ten + 1;
}
}

//handle the X
int ten = s.find_first_of('X');
if (ten >= index){
if (ten == index && s[index + 1] == 'C'){
re += 90;
index += 2;
}
else if (ten == index){
while (index < len && s[index] == 'X'){
re += 10; ++index;
}
ten = s.find_last_of('X');//"IX"
if (ten > index){
re += 9;
index = ten + 1;
}
if(index == len) return re;
}
else {
re += 9;
index = ten + 1;
}
}

//handle 'V'
int five = s.find_first_of('V');
if (five >= index){
if(five == index){
re += 5;
index++;
}
else{
re += 4;
index = five + 1;
}
}

int ge = s.find_first_of('I');
if(index > len - 1) return re;

if(ge < len - 1 && s[ge + 1] == 'X')
{
re += 9;
return re;
}
while (ge != -1 && ge <= len - 1)
{ re += 1; ge++; }
return re;
}
};