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  • longest palindromic substring

    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]

      

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  • 原文地址:https://www.cnblogs.com/whatyouknow123/p/6677615.html
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