打今儿起,小明酱说话要用京腔嘞,瞧好吧您~
嘿嘿!
原题:
Longest Palindromic Substring:
Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.
Input: "babad"
Output: "bab"
Note: “aba” is also a valid answer.
another:
Input: "cbbd"
Output: "bb"
思路:
枚举回文串的中点,这里要分为两种情况:
一种是回文串长度是奇数的情况,另一种是回文串长度是偶数的情况,枚举中点再判断是否是回文串,这样能把算法的时间复杂度降为O(n^2),但是当n比较大的时候仍然无法令人满意。
代码:
class Solution:
def longestPalindrome(self, s):
"""
:type s: str
:rtype: str
"""
res = ""
for i in range(len(s)):
res = max(self.helper(s,i,i), self.helper(s,i,i+1), res, key=len)
return res
def helper(self,s,l,r):
while 0<=l and r < len(s) and s[l]==s[r]:
l-=1; r+=1
return s[l+1:r]
结果:
Runtime: 1072 ms
好慢哦~还不是因为你太菜!哼~
大佬请喝茶(其实是整理的非常好的博文)