几百个测试用例一致通过

参考链接：https://www.zhihu.com/question/373781465/answer/1034511974命令行装了个MySQLdb，但是PyCharm似乎不识别。去file->setting->interpreter看看，发现project下有这个包：根据知乎大佬指引，需要修改下项目配置文件。打开venv下的pyvenv.cfg文件...

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.合并k个有序链表。最直接的想法先合并两条链表，然后再与第三条链表进行合并，那么时间复杂度是o(2n+3n……kn)=o(nk²)另一种想法是借鉴归并排序的思想，两两合并。T(n)=2T(n

A binary watch has 4 LEDs on the top which represent the hours (0-11), and the 6 LEDs on the bottom represent the minutes (0-59).Each LED represents a zero or one, with the least significant bit o

Given an array of integers and an integer k, you need to find the number of unique k-diff pairs in the array. Here a k-diff pair is defined as an integer pair (i, j), where i and j are both numbers in

这个题和Best Time to Buy and Sell Stock 是有关的。先看下Best Time to Buy and Sell Stock II：Say you have an array for which the ith element is the price of a given stock on day i.If you were only permitt

Given two integer arrays A and B, return the maximum length of an subarray that appears in both arrays.Example 1:Input:A: [1,2,3,2,1]B: [3,2,1,4,7]Output: 3Explanation: The repeated sub

Given two integers n and k, you need to construct a list which contains n different positive integers ranging from 1 to n and obeys the following requirement: Suppose this list is [a1, a2, a3,

Given a char array representing tasks CPU need to do. It contains capital letters A to Z where different letters represent different tasks.Tasks could be done without original order. Each task could b...

In a given array nums of positive integers, find three non-overlapping subarrays with maximum sum.Each subarray will be of size k, and we want to maximize the sum of all 3*k entries.Return the result ...

Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).Find the minimum element.You may assume no duplicate exis...

Given a range [m, n] where 0 For example, given the range [5, 7], you should return 4.这个题开始想了想，以为直接loop就行了，但是又一想好歹是个中等难度，有这么简单？管他呢就简单loop。果然==Time Limit Exceeded。那有什么骚操作呢？就参考了别人的解法。有这么一个思路。说结果

Given a string and an integer k, you need to reverse the first k characters for every 2k characters counting from the start of the string. If there are less than k characters left, reverse all of th

for遍历字典：d={'x':1,'y':2,'z':3}for k,v in d.items(): print k,v并行迭代：a=[1,2,3]b=[4,5,6]for x,y in zip(a,b): print (x,y)zip是把list压缩到一起enumerate。对于一个可迭代的（iterable）/可遍历的对象（如列表、字符串），enumerate将其组成一个索引序列

文档化函数：def f(): 'print hello' print "hello"print f.__doc__

python3普通除法：1/2=0.5（python2 是0）整除：1//2=0十六进制表示：0xAf（175）八进制表示：0o10(0o开头，十进制的8）获取输入：x=input("x:")raw_inpurt()#string四舍五入：roundround的坑：python2中round(0.5)=1,python3中round(0.5)=0.p

字符串不可变，因此不可分片赋值转换说明符%：>>> a='a%sb'>>> a%'b''abb'>>> find:子串查找：>>> a='fdfadsafd'>>> a.find('fd')0>>> 连接序列中的元素：join：>>> a=['1','2','3']>>> '.'.join(a)'1.2.3'#必须是str不能是int替换单个字符 translate

列表可修改，元组不可修改，因此列表不可做字典的键序列（列表和元组）、映射（字典）都属于容器通用序列操作：索引，分片，加，乘，判断元素是否属于序列等删除>>> name=['a','b','b']>>> del name[2]>>> name['a', 'b']分片：>>> list("naive")['n', 'a', 'i', 'v', 'e']append：在列表尾部加元素co

需求：有这样一个网站：http://www.5er0.com/.网站可以搜索有关电影电视剧有关的信息或者下载链接。现在要输入一个video name，爬到其下载链接。首先我们打开网站首页：看到有个搜索框，我们尝试输入战狼并点击搜索，得到结果如下：既然爬虫我们就要看一下数据是如何请求的，回到首页，审查元素查看搜索框的元素：看到这是一个post表单。再看搜索结果页...

堆排序的Python实现

Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times.You may assume that the array is non-empty and the majority element

Given an array of integers where 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once.Find all the elements of [1, n] inclusive that do not appear in this array.

关键字方式创建字典d=dict(name='tom',age=20)d{'age': 20, 'name': 'tom'}key必须是不可变类型，数字字符串元组字典格式化字符串d={'a':123}print '%(a)d'%d123copy:返回一个相同键值对的新字典，但是这个是浅拷贝。如果对副本值进行替换，原字典不受影响：a=dict(name='jzm',age=90)b=a.co