【hot100】刷题记录(40)-最长回文子串

题目描述：

给你一个字符串 s，找到 s 中最长的 回文 子串。

 

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"

 

提示：

• 1 <= s.length <= 1000
• s 仅由数字和英文字母组成

 

我的作答：

动态规划。。

class Solution:
    def longestPalindrome(self, s: str) -> str:
        if not s: return ''
        n = len(s)
        if n<2: return s
        max_len = 1
        begin = 0
        dp = [[False]*n for _ in range(n)]
        for i in range(n):
            dp[i][i] = True
        
        for L in range(2, n+1):
            for i in range(n):
                j = L+i-1
                if j>=n: break
                if s[i]!=s[j]:
                    dp[i][j] = False
                else:
                    if j-i<3: dp[i][j] = True
                    else: dp[i][j] = dp[i+1][j-1] 
                if dp[i][j] and j-i+1>max_len: #如果子串回文且两头元素相等
                    max_len = j-i+1
                    begin = i
        return s[begin:begin+max_len]

 

参考：

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]