needs to be changed with the if loop as in the original explanation
if ((i G.T. palLength)&&(str.charAt(i - palLength - 1) == str.charAt(i)))
{
palLength += 2;
i += 1;
continue;
}

otherwise you end up getting an out of bounds exception for some test cases like ‘aabbbb’

let me know if my interpretation of the java/c++ code is incorrect

int fastFindPalindrome(string &str)
{
// Adaptation by Artem Grotov
// Further adaptated by Mokhtar M. Khorshid

int strLength = str.length();
vector vec;
int i = 0;
int palLength = 0;

int Longest = 0;
while(i palLength) && (str[i - palLength - 1] == str[i]) )
{
palLength += 2;
i += 1;
}
//we expanded, push the length to the length array
vec.push_back(palLength);
Longest = max (Longest, palLength);
//now loop through the prefix of the polygon to see the
int s = vec.size() -2;
int e = s - palLength;
//see if we found a palindrome prefix sharing the left edge with the current palindrome
bool found = false;
for(int j = s; j > e; –j)
{
int d = j -e -1;
if(vec[j]==d)
{
palLength = d;
found = true;
break;
}
vec.push_back(min(d, vec[j]));
Longest = max (Longest, min (d, vec[j]));
}
if(!found)
{
//polygon at i+1 should be a letter rather than a space between so palLength = 1 at least
i+=1;
palLength = 1;
}
}
vec.push_back(palLength);
Longest = max (Longest, palLength);
//now we can fill in the loop if we want to get all the palindrome centers, but we don’t need it
//now we can scan through the array to find the biggest value and deduct the string of the longest palindrome

return Longest;
}

For each center, to find whether it is a palindrome, it take O(n) and it’s possible we use each letter in string as center so there is n centers.

so the run-time is o(n^2)?

I know I am missing something in the run-time analysis. Can you go into details on why the run-time is O(n)?

}

And I don’t even read 4chan. >:(

I didn’t know you had a blog. I expected more 4chan-like somethingawfulness from you.