Leetcode Problem Rating Project Introduction This frontend project has been deployed to Github Pages. The corresponding backend calculates the rating of the problems in leetcode weekly/biweekly contests based on Elo rating system as well as Maximum likelihood estimation. It is not a 100% accurate ...
Leetcode Problem Rating Project Introduction This frontend project has been deployed toGithub Pages. The corresponding backend calculates the rating of the problems in leetcode weekly/biweekly contests based onElo rating systemas well asMaximum likelihood estimation. It is not a 100% accurate result,...
题目链接 : leetcode-cn.com/problem 题目描述: 老师想给孩子们分发糖果,有 N 个孩子站成了一条直线,老师会根据每个孩子的表现,预先给他们评分。 你需要按照以下要求,帮助老师给这些孩子分发糖果: 每个孩子至少分配到 1 个糖果。 相邻的孩子中,评分高的孩子必须获得更多的糖果。 那么这样下来,老师至少需要准备多...
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只有上problem rating的那个网站(我主页贴过)其实才能回答,虽然这个量化指标不算unbais但比起多数人在借题发挥也没有更好的评价方式了,依然还是只能去借助OJ的标准去benchmark这些东西,你脱离了OJ语境这问题就变成不可定义不可描述的了,那问起来还有意义吗?纯情绪发泄了。问题下有个回答就是在批判ACM的,他的...
大餐注意，只要餐品下标不同，就可以认为是不同的餐品，即便它们的美味程度相同。两数之和问题，记得考虑溢出问题！我
something need to attention, the requirement of this problem is that only children with a higher rating need to get more candies than their neighors, but if two child has the same rating, we do not need to give more candies the time complexity is O(n), the space complexity is also O...
Timer:Timer, when it is turned on, it will prompt the problem solving time in the status bar in the lower right corner Editor Menu(Right-click on Editor to appear):Function as above Question Editor: Content:Show content(Rely on Markdown) Solution:show solution Submissions:View the submissio...
Children with a higher rating get more candies than their neighbors. What is the minimum candies you must give? 【解答】最开始我选择的做法是像分析函数曲线一样找拐点,如果知道这条曲线的拐点,比如极大值,那么极大值左侧是递增的,极大值右侧是递减的;极小值左侧递减,右侧递增。这里特别麻烦的地方在于,...
Contest Rating 1,789 Global Ranking 54,583/697,019 Attended 6 -/- Solved --% Acceptance Beats --% Beats --% Beats --% - Attempting - submission - Attempting - Attempting - Attempting Badges 0 Locked Badge May LeetCoding Challenge List Solutions DiscussCopyright...