#其中ham没有默认值,而eggs是由默认值的,其默认值为'spam'.#参数ham的注释部分为:42;参数eggs的注释部分为:int#"Nothing to see here"是返回值的注释,这个必须用 '->'连接#看了java代码后,你会有更直观的认识,注释嘛,你可以根据你自己的想法,想怎么写就怎么写,如42,int;不过不好的注释有时候会给别人阅...
python Solution LeetCode No.409 Longest Palindrome(最长回文串) 题目: 给定一个包含大写字母和小写字母的字符串,找到通过这些字母构造成的最长的回文串。 在构造过程中,请注意区分大小写。比如 "Aa" 不能当做一个回文字符串。 注意: 假设字符串的长度不会超过 1010。 示例 1: 输入: "abccccdd" 输出: 7...
Python3代码 classSolution:deflongestPalindrome(self, s:str) ->int:importcollections# 统计各字符个数count = collections.Counter(s).values()sum=0forxincount:ifx //2>0:# 取偶数个字符sum+= x //2*2ifsum==len(s):returnsumelse:returnsum+1 代码地址 GitHub链接...
class Solution: def longestPalindrome(self, s): """ :type s: str :rtype: str """ l = len(s) max_length = 0 palindromic = '' for i in range(l): x = 1 while (i - x) >= 0 and (i + x) < l: if s[i + x] == s[i - x]: x += 1 else: break x -= 1 if...
class Solution: def longestPalindrome(self, s: str) -> str: #def longestPalindrome(s: str) -> str: n = len(s) if n < 2: return s # 如果字符串长度小于2,它本身就是最长的回文子串 # dp[i][j]表示s[i:j+1]是否是回文串 dp = [[False] * n for _ in range(n)] start, max_...
技术标签: python leetcodeTitle: Given a string which consists of lowercase or uppercase letters, find the length of the longest palindromes that can be built with those letters. This is case sensitive, for example “Aa” is not considered a palindrome here. Note: Assume the length of ...
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] ...
letterthatwas added in t.题目14:409.LongestPalindromeGivenastringwhichconsistsoflowercaseoruppercaseletters,findthelengthofthelongestpalindromesthatcanbebuiltwiththoseletters. EDX Python WEEK 1 problem 3 Longest substring_ Answer Writeaprogramthatprintsthelongestsubstringofs inwhichthelettersoccur in alphabetica...
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 - ...
Python代码: classSolution(object):deflongestPalindrome(self, s):""" :type s: str :rtype: int """ans = odd =0cnt = collections.Counter(s)forcincnt: ans += cnt[c]ifcnt[c] %2==1: ans -=1odd +=1returnans + (odd >0) ...