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-1
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