本文实例讲述了Python3最长回文子串算法。分享给大家供大家参考，具体如下：
1. 暴力法
思路：对每一个子串判断是否回文
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class Solution: def longestPalindrome(self, s): """ :type s: str :rtype: str """ if len(s) == 1: return s re = s[0] for i in range(0,len(s)-1): for j in range(i+1,len(s)): sta = i end = j flag = True while sta < end: if s[sta] != s[end]: flag = False break sta += 1 end -= 1 if flag and j-i+1 > len(re): re = s[i:j+1] return re |
提交结果：超出时间限制
2. 动态规划法
思路：
m[i][j]标记从第i个字符到第j个字符构成的子串是否回文，若回文值为True，否则为False.
初始状态 s[i][i] == True，其余值为False.
当 s[i] == s[j] and m[i+1][j-1] == True 时，m[i][j] = True
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class Solution: def longestPalindrome(self, s): """ :type s: str :rtype: str """ k = len(s) matrix = [[False for i in range(k)] for j in range(k)] re = s[0:1] for i in range(k): for j in range(k): if i==j: matrix[i][j] = True for t in range(1,len(s)): #分别考虑长度为2~len-1的子串（长串依赖短串的二维数组值） for i in range(k): j = i+t if j >= k: break if i+1 <= j-1 and matrix[i+1][j-1]==True and s[i] == s[j]: matrix[i][j] = True if t+1 > len(re): re = s[i:j+1] elif i+1 == j and j-1 == i and s[i] == s[j]: matrix[i][j] = True if t+1 > len(re): re = s[i:j+1] return re |
执行用时：8612 ms
希望本文所述对大家Python程序设计有所帮助。
原文链接：https://blog.csdn.net/aawsdasd/article/details/80484938
