Python3最长回文子串算法 Python3最长回文子串算法示例

时间:2021-03-30 gxnustc 人气:0

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

希望本文所述对大家Python程序设计有所帮助。

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