#-*- coding: UTF-8 -*- class Solution(object):    def isPalindrome(self, s):        """        :type s: str        :rtype: bool        """        s=s.lower()        if s==None:return False        isPalindrome1=[]        isPal…

#-*- coding: UTF-8 -*-#利用栈的思想#如果输入的左测扩则入栈,如果输入为右侧扩,判断其是否与当前栈顶配对,如果匹配则删除这一对,否则return False#'(', ')', '{', '}', '[' and ']'class Solution(object):    def isValid(self, s):        """        :type s: str        :rtype: bool        ""…

#-*- coding: UTF-8 -*-#特定的九个格内1-9的个数至多为1#依次检查每行,每列,每个子九宫格是否出现重复元素,如果出现返回false,否则返回true.class Solution(object):    def isValidSudoku(self, board):        """        :type board: List[List[str]]        :rtype: bool        """  …

#-*- coding: UTF-8 -*- from collections import Counterclass Solution(object):    def longestPalindrome(self, s):        """        :type s: str        :rtype: int        """        lopa=0        lopa1=0        slist=list(s)  …

class Solution(object):    def isAnagram(self, s, t):        if sorted(list(s.lower()))==sorted(list(t.lower())):            return True        return Falsesol=Solution()print sol.isAnagram(s='anAgram', t='nagaram')…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 解题方法 列表生成式 正则表达式 双指针 日期 题目地址:https://leetcode.com/problems/valid-palindrome/description/ 题目描述 Given a string, determine if it is a palindrome, considering only alphanumeric characters an…

一天一道LeetCode 本系列文章已全部上传至我的github,地址:ZeeCoder's Github 欢迎大家关注我的新浪微博,我的新浪微博 欢迎转载,转载请注明出处 (一)题目 Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases. For example, "A man, a plan, a canal: Panama&qu…

题目: Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases. For example,"A man, a plan, a canal: Panama" is a palindrome."race a car" is not a palindrome. Note:Have you consider tha…

#-*- coding: UTF-8 -*- #既然不能使用加法和减法,那么就用位操作.下面以计算5+4的例子说明如何用位操作实现加法:#1. 用二进制表示两个加数,a=5=0101,b=4=0100:#2. 用and(&)操作得到所有位上的进位carry=0100;#3. 用xor(^)操作找到a和b不同的位,赋值给a,a=0001:#4. 将进位carry左移一位,赋值给b,b=1000:#5. 循环直到进位carry为0,此时得到a=1001,即最后的sum.#!!!!!!关于负数的运算.…

#-*- coding: UTF-8 -*-class Solution(object):    def isPalindrome(self, head):        """        :type head: ListNode        :rtype: bool        """        if head==None or head.next==None:return True        resList=[]       …