s = Solution() string = "cubwqvhxdammpkwkycrqtegepyxtohspeasrdtinjhbesilsvffmnzznmltsspjwuogdyzvanalohmzrywdwqqcukjceothydlgtocukc" #string = "abbabcba" print s.longestPalindrome(string) if __name__ == "__main__": import time start = time.clock() main() print "%s sec" % (tim...
题意:Given a string S, find the longest palindromic substring in S. You may assume that the maximum length of S is 1000, and there exists one unique longest palindromic substring.解题思路:最长回文子串求解。代码:class Solution: # @return a string def getlongestpalindrome(self, s, l, r): ...
classSolution:deflongestPalindrome(self,s):""":type s:str:rtype:str""" l=len(s)max_length=0palindromic=''iflen(s)==1:returnsforiinrange(l):forjinrange(i+1,l):is_palindromic=Trueforkinrange(i,int((i+j)/2)+1):ifs[k]!=s[j-k+i]:is_palindromic=Falsebreakifis_palindromicand(...
空间复杂度O(1) deflongestPalindrome(self,s):iflen(s)<2:returns start=maxlen=0foriinrange(1,len(s)):start,maxlen=self.Palindrome(s,i-1,i+1,maxlen,start)start,maxlen=self.Palindrome(s,i-1,i,maxlen,start)returns[start:start+maxlen]defPalindrome(self,s,left,right,maxlen,start):whi...
def longestPalindrome(self, s): """ :type s: str :rtype: str """ if not s: return '' # 先将任一单个字母作为结果 start = 0 maxlen = 1 dp = [[0 for __ in range(len(s))] for __ in range(len(s))] # 将长度1和长度2(相同字母)情况初始化赋值 ...
palindrome=''foriinrange(len(s)):len1=len(self.getlongestpalindrome(s,i,i))# len1是中间是单一字母的回文数iflen1>len(palindrome):palindrome=self.getlongestpalindrome(s,i,i)len2=len(self.getlongestpalindrome(s,i,i+1))# len2是中间是和后面一个字母相同的回文数iflen2>len(palindrome):pali...
def longestPalindrome(self, s): res = "" for i in xrange(len(s)): # odd case, like "aba" tmp = self.helper(s, i, i) if len(tmp) > len(res): res = tmp # even case, like "abba" tmp = self.helper(s, i, i+1) ...
给定一个字符串s,找到s中最长的回文子串。你可以假设s的最大长度为 1000 代码:(参考:https://leetcode.com/problems/longest-palindromic-substring/discuss/3337/Manacher-algorithm-in-Python-O(n)) classSolution(object):deflongestPalindrome(self,s):""" ...
classSolution:deflongestPalindrome(self,s:str)->str:n=len(s)ifn<2:returnsimportnumpyasnpp=np.zeros((n,n))foriinrange(n):p[i,i]=1ss=[0,0]# p(i,i) == Trueforiinrange(1,n):j=i+1i=i-1whilei>=0andj<=n-1:ifs[i]==s[j]andp[i+1,j-1]:p[i,j]=1# num = j - ...
def longestPalindrome(self, s: str) -> str: l, ms = len(s), s[0] if s else '' # 内循环少一次 if l < 2: return s for i in range(l): for j in range(i + len(ms), l):# 关键点一: + len(ms) x = s[i:j + 1] ...