有两个最大长度的子字符串时如何找到最长的回文子字符串

Question

I am solving the longest palindrome substring problem from here . I am stuck on how should I return the substring which occurs first ( with the least starting index ) when there is a conflict.

def longestPalindrome(A):
    pal_string = ''
    x = len(A)
    y = 0 
    for i in range(0,x):
        for j in range(x,i-1,-1):
            new_str =  A[i:j]
            if new_str == new_str[::-1]:
                if len(new_str) >= y:
                    y = len(new_str)
                    pal_string = new_str

    return pal_string
print longestPalindrome('abb')
print longestPalindrome('aaaabbaaa')
print longestPalindrome('caba')
print longestPalindrome("abbcccbbbcaaccbababcbcabca")

Input : ("abbcccbbbcaaccbababcbcabca") should return bbcccbb but my code returns cbababc .

I am using Python2.7.

Answer 1

As @Aran-Fey pointed out, you need to change

if len(new_str) >= y:

to

if len(new_str) > y:

Explanation: If you use the >= operator, then later palindromes with the same length will override earlier ones, while the > operator ensures that you will only set pal_string and y if you have found a longer palindrome.