"""
给你一个字符串 s，找到 s 中最长的回文子串。
如果字符串的反序与原始字符串相同，则该字符串称为回文字符串。
示例 1：
输入：s = "babad"
输出："bab"
解释："aba" 同样是符合题意的答案。
示例 2：
输入：s = "cbbd"
输出："bb"
"""
def longest_par(s):
if not s:
return ""
n = len(s)
dp = [[False]*n for _ in range(n)]
ans = s[0]
for i in range(n-1, -1, -1):
dp[i][i] = True
if i > 0 and s[i] == s[i-1]:
dp[i-1][i] = True
if len(ans) < 2:
ans = s[i-1:i+1]
for j in range(i+2, n):
if s[i] == s[j]:
dp[i][j] = dp[i+1][j-1]
if dp[i][j] and len(ans) < j-i+1:
ans = s[i:j+1]
return ans
print(longest_par("aabba"))
print(longest_par("aabbaa"))
print(longest_par("jbba"))
print(longest_par("jbbaa"))
print(longest_par("jbbjaa"))
print(longest_par(""))
# 下面是GPT4给的写法，感觉逻辑上比我的更清晰
def longestPalindromeDP(s):
n = len(s)
if n == 0:
return ""
start = 0
max_len = 1
dp = [[False] * n for _ in range(n)]
# 单个字符是回文
for i in range(n):
dp[i][i] = True
# 两个字符相同是回文
for i in range(n - 1):
if s[i] == s[i + 1]:
dp[i][i + 1] = True
start = i
max_len = 2
# 遍历长度大于2的子串
for length in range(3, n + 1):
for i in range(n - length + 1):
j = i + length - 1
if s[i] == s[j] and dp[i + 1][j - 1]:
dp[i][j] = True
start = i
max_len = length
return s[start:start + max_len]
# 测试示例
s1 = "babad"
print(longestPalindromeDP(s1)) # 输出："aba"
s2 = "jbbjaa"
print(longestPalindromeDP(s2))
