5. 最长回文子串
#动态规划，主要利用如果i+1到j-1是回文字符串且s[i]==s[j]那么i 到 j也是回文字符串
1 class Solution:
2 def longestPalindrome(self, s):
3 """
4 :type s: str
5 :rtype: str
6 """
7 n=len(s)
8 aa=[0]*(n*n)
9 longest=''
10 for i in range(n):
11 aa[i*n+i]=1 #单个字符肯定是回文字符串，用于回文长度为3的判断
12 longest=s[i]
13 for i in range(n-1):
14 if s[i]==s[i+1]:
15 aa[i*n+i+1]=1 #如果两个字符相同，那么它们也构成回文字符串，用于回文长度为4的判断
16 longest=s[i:i+2]
17 for l in range(3,n+1):
18 for i in range(0,n-l+1):
19 e=l+i-1
20 if (aa[(i+1)*n+e-1] & int(s[i]==s[e])):
21 longest=s[i:e+1]
22 aa[i*n+e]=1
23 return longest
24
