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
- 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
- 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
更多关于Python相关内容感兴趣的读者可查看jb51专题:《Python数据结构与算法教程》、《Python加密解密算法与技巧总结》、《Python编码操作技巧总结》、《Python函数使用技巧总结》、《Python字符串操作技巧汇总》及《Python入门与进阶经典教程》
希望本文所述对大家Python程序设计有所帮助。