#-*- coding: UTF-8 -*-class Solution(object):      def getHint(self, secret, guess):          """         :type secret: str         :type guess: str         :rtype: str         """          countA,i,numList=0,-1,[[0,0] for j…

class Solution(object):    def getHint(self, secret, guess):        """        :type secret: str        :type guess: str        :rtype: str        """        countA,i,numList=0,-1,[[0,0] for i in range(10)]        while True:…

[LeetCode]299. Bulls and Cows 解题报告(Python) 作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 题目地址:https://leetcode.com/problems/bulls-and-cows/description/ 题目描述: You are playing the following Bulls and Cows game with your friend: You write down…

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 You are playing the following Bulls and Cows game with your friend: You write down a number and ask your friend to guess what the number is. Each time your >…

#-*- 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.…

题目如下: 解题思路:本题难度不太大,对时间复杂度也没有很高的要求.我的做法是用一个字典来保存每个字符出现的次数,用正数1记录标记secret中出现的字符,用负数1记录guess中出现的字符,这样每出现一次正负抵消,即表示出现了一次cow. 代码如下: class Solution(object): def getHint(self, secret, guess): """ :type secret: str :type guess: str :rtype: str &quo…

#-*- coding: UTF-8 -*- #AC源码[意外惊喜,还以为会超时]class Solution(object):    def twoSum(self, nums, target):        """        :type nums: List[int]        :type target: int        :rtype: List[int]        """         for i in xrange(…

#-*- coding: UTF-8 -*-#利用strip函数去掉字符串去除空格(其实是去除两边[左边和右边]空格)#利用split分离字符串成列表class Solution(object):    def lengthOfLastWord(self, s):        """        :type s: str        :rtype: int        """        if s==None:return 0     …

#-*- coding: UTF-8 -*-#需要考虑多种情况#以下几种是可以返回的数值#1.以0开头的字符串,如01201215#2.以正负号开头的字符串,如'+121215':'-1215489'#3.1和2和空格混合形式[顺序只能是正负号-0,空格位置可以随意]的:'+00121515'#4.正数小于2147483647,负数大于-2147483648的数字#其他的情况都是返回0,因此在判断 是把上述可能出现的情况列出来,其他的返回0#AC源码如下class Solution(object…

#-*- coding: UTF-8 -*-class Solution(object):    def compareVersion(self, version1, version2):        """        :type version1: str        :type version2: str        :rtype: int        """        versionl1=version1.split('.'…