标签：substring sam a long self center order object code logs
problem description:
given a string s, you shou find the longest palindromic substring in there
for example:
input :"ssaass"
ouput:"ssaass"
one solution: times o(n^2):
class Solution(object): def longestPalindrome(self, s): """ :type s: str :rtype: str """ length = len(s) i = length if length == 0: return while i >1: j = 0 while j <= (length - i): s1 = s[j:j+i] s2 = s1[::-1] if s1 == s2: return s1 j += 1 i -= 1 return s[0]
but in the leedcode online judge, it take a long time to exceed, so it is not good enough.Then i write anothe code according to someone write in the discussion.It‘s main thought is that
the same words(the one word also can be explainede as the same words) should be included in the center of the substring, then you can expand this substring.There should be three point. Defining i in order to limit the position of the center, it‘s length is form 0 to len-1.Defining j, k in order to record the bounds of the palindromic substring.Maybe i don‘t explain it well, so there is my code.It is written by python.
class Solution(object): def longestPalindrome(self, s): """ :type s: str :rtype: str """ length = len(s) i = 0 minstart = 0 maxlen = 1 if length <= 1: return s[0] while i< length: while (length - i)<maxlen/2: return s[minstart:minstart+maxlen] j = i k = i while k<(length -1) and s[k]==s[k+1]: k += 1 i = k+1 while k<(length -1) and j and s[j-1]==s[k+1]: j -= 1 k += 1 if (k - j +1)>maxlen: minstart = j maxlen = k-j+1 return s[minstart:minstart+maxlen]
标签：substring sam a long self center order object code logs
原文地址：http://www.cnblogs.com/whatyouknow123/p/6677615.html
