Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.
Example 1:
Input: "babad" Output: "bab" Note: "aba" is also a valid answer.
Example 2:
Input: "cbbd" Output: "bb"
第一次：想法是将得到s的倒序字符串s1，将s与s1对应位置的元素进行比较，并保留最大长度相同的元素。这种想法是错误的！如：cbbd与dbbc，虽然可得到bb，但对于dabab与babad而言，只能得到aba 这一种情况。
"""
第二次：遍历s中的每个字符，以该字符为中心，寻找其最长回文字符串，对于长度为奇数的可行，为偶数的会出错，如cbbd是找不到bb的，解决方法是在字符与字符之间插入辅助字符‘#’，包括初始位置与末尾，这样均能保证字符串长度为奇数。
"""
class Solution(object):
def longestPalindrome(self, s):
"""
:type s: str
:rtype: str
"""
if len(s) == 1 or len(s) == 0: return s
s = '#' + '#'.join(s) + '#'
print(s)
ll = len(s)
maxLen = 0
result = []
for i in range(0, ll):
subStr = ""
for j in range(1, i + 1):
if i + j < ll and s[i - j] == s[i + j]:
pass
else:
subStr = s[i-j+1: i+j]
break
subStr = s[0:2*i+1]
if maxLen <= len(subStr):
maxLen = len(subStr)
result = subStr
return result.replace('#', '')"""
Runtime: 3080 ms, faster than 30.83% of Python online submissions forLongest Palindromic Substring.
Memory Usage: 11.7 MB, less than 29.57% of Python online submissions for Longest Palindromic Substring.
"""