题目:

题解:
class Solution:
def expand(self, s, left, right):
while left >= 0 and right < len(s) and s[left] == s[right]:
left -= 1
right += 1
return (right - left - 2) // 2
def longestPalindrome(self, s: str) -> str:
end, start = -1, 0
s = '#' + '#'.join(list(s)) + '#'
arm_len = []
right = -1
j = -1
for i in range(len(s)):
if right >= i:
i_sym = 2 * j - i
min_arm_len = min(arm_len[i_sym], right - i)
cur_arm_len = self.expand(s, i - min_arm_len, i + min_arm_len)
else:
cur_arm_len = self.expand(s, i, i)
arm_len.append(cur_arm_len)
if i + cur_arm_len > right:
j = i
right = i + cur_arm_len
if 2 * cur_arm_len + 1 > end - start:
start = i - cur_arm_len
end = i + cur_arm_len
return s[start+1:end+1:2]
本文介绍了如何通过Solution类中的expand函数辅助计算字符串中两个指针之间的最长相同子串,然后在longestPalindrome函数中应用此方法寻找给定字符串的最长回文子串。

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