O(n^2) solution to GPCv1 Level 1

v1_level1_n_2.py

def get_longest_palindrome(text):
    best_palindrome = ''
    length = len(text)
    for i in range(length * 2 - 1):
        start = i / 2 - i % 2
        end = i / 2 + 1
        while start > 0 and end < length and text[start] == text[end]:
            start -= 1
            end += 1
        candidate = text[start+1:end]
        if len(candidate) > len(best_palindrome):
            best_palindrome = candidate
    return best_palindrome