#-*- coding: UTF-8 -*- #l1 = ['1','3','2','3','2','1','1']#l2 = sorted(sorted(set(l1),key=l1.index,reverse=False),reverse=True)class Solution(object):    def thirdMax(self, nums):        """        :type nums: List[int]        :rtype: int  …

#-*- 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.…

problem 414. Third Maximum Number solution 思路:用三个变量first, second, third来分别保存第一大.第二大和第三大的数,然后遍历数组. class Solution { public: int thirdMax(vector<int>& nums) { //1.LONG_MIN; long first = LONG_MIN, second = LONG_MIN, third = LONG_MIN;// for(auto a:n…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 替换最大值数组 使用set 三个变量 日期 题目地址:https://leetcode.com/problems/third-maximum-number/description/ 题目描述 Given a non-empty array of integers, return the third maximum number in this arr…

Description Given a non-empty array of integers, return the third maximum number in this array. If it does not exist, return the maximum number. The time complexity must be in O(n). Example 1: Input: [, , ] Output: Explanation: The third maximum .  E…

#-*- coding: UTF-8 -*-# The guess API is already defined for you.# @param num, your guess# @return -1 if my number is lower, 1 if my number is higher, otherwise return 0# def guess(num):#binary searchclass Solution(object):    def guessNumber(self, n…

#回文数#Method1:将整数转置和原数比较,一样就是回文数:负数不是回文数#这里反转整数时不需要考虑溢出,但不代表如果是C/C++等语言也不需要考虑class Solution(object):    def isPalindrome(self, x):        """        :type x: int        :rtype: bool        """        if x<0:return False    …

#-*- coding: UTF-8 -*- # Definition for a binary tree node.# class TreeNode(object):#     def __init__(self, x):#         self.val = x#         self.left = None#         self.right = Noneclass Solution(object):    def maxDepth(self, root):        if…

#-*- coding: UTF-8 -*- class Solution(object):    hexDic={0:'0',1:'1',2:'2',3:'3',4:'4',5:'5',6:'6',7:'7',8:'8',9:'9',\            10:'a',            11:'b',            12:'c',            13:'d',            14:'e',            15:'f'}            def t…

class Solution(object):    def isUgly(self, num):        if num<=0:return False        comlist=[2,3,5];modflag=0                while True:            if num==1:break            for value in comlist:                tuple=divmod(num,value)            …