# 给定一个字符串 s,找到 s 中最长的回文子串。你可以假设 s 的最大长度为 1000。
#
# 示例 1:
#
# 输入: "babad"
# 输出: "bab"
# 注意: "aba" 也是一个有效答案。
#
#
# 示例 2:
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# 输入: "cbbd"
# 输出: "bb"
#
# Related Topics 字符串 动态规划
# 👍 2558 👎 0
# leetcode submit region begin(Prohibit modification and deletion)
class Solution(object):
def longestPalindrome(self, s):
"""
:type s: str
:rtype: str
"""
max_l = 0
res = ""
for i in range(0, len(s)):
# s[i] 为中心
left, right = i, i
while left >= 0 and right < len(s) and s[left] == s[right]:
if max_l < right - left + 1:
max_l = right - left + 1
res = s[left:right + 1]
left -= 1
right += 1
# s[i] s[i+1] 为中心
left, right = i, i + 1
while left >= 0 and right < len(s) and s[left] == s[right]:
if max_l < right - left + 1:
max_l = right - left + 1
res = s[left:right + 1]
left -= 1
right += 1
return res
# leetcode submit region end(Prohibit modification and deletion)
s = Solution()
print(s.longestPalindrome("babad"))
print(s.longestPalindrome("cbbd"))
class Solution(object):
def longestPalindrome(self, s):
"""
:type s: str
:rtype: str
"""
start, end = 0, 0
for index, item in enumerate(s):
left1, right1 = self.expandAroundCenter(s, index, index)
left2, right2 = self.expandAroundCenter(s, index, index + 1)
if right1 - left1 > end - start:
start, end = left1, right1
if right2 - left2 > end - start:
start, end = left2, right2
return s[start: end + 1]
def expandAroundCenter(self, s, left, right):
while left > 0 and right < len(s) and s[left] == s[right]:
left -= 1
right += 1
return left + 1, right - 1