[leetcode] 5.Longest Palindromic Substring

开始觉得挺简单的写完发现这个时间超限了：

class Solution:
    def longestPalindrome(self, s: str) -> str:
        # longest palindromic substring lenth < 1000
        #NULL
        string = s
        if len(string) == 0:
            return ''
        # ---
        # regular
        # a not empty string must have 1 letterd palindromic substring:string[0]
        longestPalindSubStringLength = 1
        longestPalindSubString = string[0]
        for start in range(len(string)):
            for end in range(start + 1, len(string) +1): #[] operator is right open interval,will not count "end" if no +1
                # one letter
                if len(string[start:end]) == 1:
                    continue
                if (self.isPalind(string[start:end])):
                    if len(string[start:end]) > longestPalindSubStringLength:# > will get bab ;>= will get aba. I prefer simple one
                        longestPalindSubStringLength = len(string[start:end])
                        longestPalindSubString = string[start:end]
        return longestPalindSubString

    def isPalind(self, substring: str) -> bool:
        #notice substring >= 2
        substringlength = len(substring)

        looplimit = substringlength // 2
        for index in range(0, looplimit):
            if substring[index] != substring[substringlength-1 -index]: # axisymmetric
                return False
            else:
                pass
        # run to here without return mean palindromic
        return True

"abababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababa"

用pycharm跑了下要4秒，结果就是它自己。[testcase:85]

发现之前判断轴对称的方法太低级了不用逐位判断轴对称的前一半和后一半的逆序是同一字符串即可。

class Solution:
    def longestPalindrome(self, s: str) -> str:
        # longest palindromic substring lenth < 1000
        #NULL
        string = s
        if len(string) == 0:
            return ''
        # ---
        # regular
        # a not empty string must have 1 letterd palindromic substring:string[0]
        longestPalindSubStringLength = 1
        longestPalindSubString = string[0]
        for start in range(len(string)):
            for end in range(start + 1, len(string) +1): #[] operator is right open interval,will not count "end" if no +1
                # one letter
                if len(string[start:end]) == 1:
                    continue
                if (self.isPalind(string[start:end])):
                    if len(string[start:end]) > longestPalindSubStringLength: #> mean bab;>= mean aba
                        longestPalindSubStringLength = len(string[start:end])
                        longestPalindSubString = string[start:end]
        return longestPalindSubString

    def isPalind(self, substring: str) -> bool:
        #notice substring >= 2
        substringlength = len(substring)

        if(substringlength %2)  == 1:
            #odd
            temp1 = substring[0:substringlength // 2 +1]
            temp2 = substring[substringlength:substringlength // 2 - 1:-1]
            if temp1 == temp2:
                return True
            else:
                return False
        else:
            #even
            temp1 = substring[0:substringlength//2]
            temp2 = substring[substringlength:substringlength//2 -1:-1]
            if temp1 == temp2:
                return True
            else:
                return False

testcase:95

"ffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggg"

f比g多一个

[leetcode] 5.Longest Palindromic Substring

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