leetcode算法题基础(三十五)动态规划(三)5. 最长回文子串
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:
输入: "cbbd"
输出: "bb"
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/longest-palindromic-substring
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
class Solution(object): def longestPalindrome(self, s): n = len(s) dp = [[0] * n for _ in range(n)] max_len = float("-inf") res = "" for i in range(n): # dp[i][i] = 1 for j in range(i, -1, -1): if s[i] == s[j] and (i - j < 2 or dp[i - 1][j + 1]): dp[i][j] = 1 if dp[i][j] and i - j + 1 > max_len: max_len = i - j + 1 res = s[j:i + 1] # print(dp) return res
关键:
循环字符串一次,以回文长度为基础,试探当前长度,当前循环位置下,
(1)前后各加一是否构成回文,(2)或者往后加一是否构成回文。
之后返回起始位置及最长长度构成的字符串
class Solution(object): def longestPalindrome(self, s): str_length = len(s) max_length = 0 start = 0 for i in range(str_length): if i - max_length >= 1 and s[i - max_length - 1:i + 1] == s[i - max_length - 1:i + 1][::-1]: start = i - max_length - 1 max_length += 2 continue if i - max_length >= 0 and s[i - max_length:i + 1] == s[i - max_length:i + 1][::-1]: start = i - max_length max_length += 1 return s[start:start + max_length]
本文来自博客园,作者:秋华,转载请注明原文链接:https://www.cnblogs.com/qiu-hua/p/14004611.html