#-*- coding: UTF-8 -*- class Solution(object):    def hammingWeight(self, n):        if n<=0:return n        mid=[]        while True:            if n==0:break            n,mod=divmod(n,2)            mid.append(mod)        mid.reverse()        return…

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 右移32次 计算末尾的1的个数 转成二进制统计1的个数 使用mask 日期 [LeetCode] 题目地址:https://leetcode.com/problems/number-of-1-bits/ Total Accepted: 88721 Total Submissions: 236174 Difficulty: Easy 题目描述 Writ…

#-*- 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.…

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 Write a function that takes an unsigned integer and returns the number of '1' bits it has (also known as the Hamming weight). For example, the 32-bit integer…

题目: Write a function that takes an unsigned integer and returns the number of ’1' bits it has (also known as the Hamming weight). For example, the 32-bit integer ’11' has binary representation 00000000000000000000000000001011, so the function should…

Number of 1 Bits Write a function that takes an unsigned integer and return the number of '1' bits it has (also known as the Hamming weight). Example 1: Input: 00000000000000000000000000001011 Output: 3 Explanation: The input binary string 0000000000…

#-*- 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…

#-*- 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  …

#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 -*- 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…