第三步:case when time1=1取出我们要的来计数;全体用户在我们取首次登陆时间的a就已经取出来了。 select round(sum(case when time1=1 then 1 else 0 end)/count(distinct player_id),2) as fraction from ( select a.player_id, datediff(b.event_date,a.first_log_date) as time1 ...
682. 棒球比赛 - 你现在是一场采用特殊赛制棒球比赛的记录员。这场比赛由若干回合组成,过去几回合的得分可能会影响以后几回合的得分。 比赛开始时,记录是空白的。你会得到一个记录操作的字符串列表 ops,其中 ops[i] 是你需要记录的第 i 项操作,ops 遵循下述规则: 1.
0511 Game Play Analysis I 80.8% Easy 0512 Game Play Analysis II 55.5% Easy 0513 Find Bottom Left Tree Value Go 61.5% Medium 0514 Freedom Trail 43.0% Hard 0515 Find Largest Value in Each Tree Row Go 61.1% Medium 0516 Longest Palindromic Subsequence 53.2% Medium 0517 Super Washing ...
2. The problem discussion is for asking questions about the problem or for sharing tips - anything except for solutions. 3. If you'd like to share your solution for feedback and ideas, please head to the solutions tab and post it there. Sort by:Best No comments yet. 123 ...
LeetCode Solutions: A Record of My Problem Solving Journey.( leetcode题解,记录自己的leetcode解题之路。) 235 soimort/you-get Python 55.01k :arrow_double_down: Dumb downloader that scrapes the web 236 Solido/awesome-flutter Dart 54.849k An awesome list that curates the best Flutter li...
CasparCG - A Windows and Linux software used to play out professional graphics, audio and video to multiple outputs as a layerbased real-time compositor DJV - Professional review software for VFX, animation, and film production ffmpeg - A complete, cross-platform solution to record, convert and...
classSolution(object): defxorGame(self, nums): """ :type nums: List[int] :rtype: bool """ xo=0 foriinnums: xo ^=i returnxo==0orlen(nums)%2==0 Analysis: Math: Corner Case: If the XOR of all nums is 0, then A wins. ...
* @lc app=leetcode id=45 lang=cpp * * [45] Jump Game II */ // @lc code=start class Solution { // assume that you can always reach the last index. public: int jump(vector<int>& nums) { assert(!nums.empty()); if (nums.size() == 1) return 0; ...
LeetCode "Flip Game II" A natural recursion thought.. Please note we can cache intermediate results. classSolution { unordered_map<string,bool>hs;public:boolcanWin(strings) {if(hs.count(s)) {returnhs[s]; } size_t n=s.length();if(n <2)returnfalse;boolret =false;inti =0;while(i ...
= 0. The explanation come from https://leetcode.com/problems/chalkboard-xor-game/discuss/165396/Detailed-math-explanation-Easy-to-understand 永远渴望,大智若愚(stay hungry, stay foolish)