我想写一个python函数来有效地做到这一点:
该函数将接受两个字符串，'a'和'b'，并试图找到最长的回文字符串 它可以被形成为使得它是'a'的非空子串和一个非空子串的级联， 子字符串'b'.如果有多个有效答案，它将返回字典中最小的一个. 如果不能形成这样的字符串，它将返回'—1'.
我有一个低效的解决方案，它生成两个字符串的所有子字符串，然后创建所有可能的串联，跟踪最长的有效回文:
def is_palindrome(word):
"""Check if a word is a palindrome."""
reversed_word = word[::-1]
return word == reversed_word
def all_substrings_of_word(word):
"""Generate all possible non-empty substrings of a given string."""
substrings = []
for sub_string_length in range(1, len(word) + 1):
for i in range(len(word) - sub_string_length + 1):
new_word = word[i:i + sub_string_length]
substrings.append(new_word)
return substrings
def buildPalindrome(a, b):
"""Attempt to find the longest palindromic string created by concatenating
a substring of `a` with a substring of `b`."""
sub_strings_a = all_substrings_of_word(a)
sub_strings_b = all_substrings_of_word(b)
# Generate all possible concatenations of substrings from `a` and `b`
multiplexed_array = [
word_a + word_b for word_a in sub_strings_a for word_b in sub_strings_b]
# Find the best palindrome (longest, then lexicographically smallest)
best_palindrome = ""
for word in multiplexed_array:
if is_palindrome(word):
if len(word) > len(best_palindrome):
best_palindrome = word
elif len(word) == len(best_palindrome) and word < best_palindrome:
best_palindrome = word
return best_palindrome if best_palindrome else "-1"
print(buildPalindrome("bac", "bac")) # EXPECTED OUTPUT -- aba
print(buildPalindrome("abc", "def")) # EXPECTED OUTPUT -- -1
print(buildPalindrome("jdfh", "fds")) # EXPECTED OUTPUT -- dfhfd
我能请你解释一下如何改进这一点吗？