https://leetcode.cn/problems/cat-and-mouse-ii/ 情况同913。但这道题更离谱的是,假设任意一方获胜都不需要2次经过相同位置,而对步数设上限的做法没有正确性,但8*8的棋盘内没有反例。所以DP做法是否算正解至今仍有争议。 Top22 3022 给定操作次数内使剩余元素的或值最小 https://leetcode.cn/problems/min...
空间复杂度: O(mn) 该算法题解的仓库:LeetCode-Swift[2] 前往LeetCode[3] 练习 参考资料 [1]@故胤道长: https://m.weibo.cn/u/1827884772 [2]LeetCode-Swift: https://github.com/soapyigu/LeetCode-Swift [3]LeetCode: https://leetcode.com/problems/regular-expression-matching...
Each row of this table indicates the ID of a department and its name. A company's executives are interested in seeing who earns the most money in each of the company's departments. Ahigh earnerin a department is an employee who has a salary in thetop three uniquesalaries for that depart...
LeetCode 301960 Jul 11, 2017 Editorial SolutionpandasApproach 1: Return the First n Rows Using nlargest()AlgorithmFor this problem, we can either identify the top earners first using DataFrame employee and then join the DataFrame department to get the department name, or join the...
Leetcode - 最大子序和 https://leetcode-cn.com/problems/maximum-subarray/submissions/ 题目描述 给定一个整数数组nums,找到一个具有最大和的连续子数组(子数组最少包含一个元素),返回其最大和。 示例: 输入:[-2,1,-3,4,-1,2,1,-5,4],输出:6解释:连续子数组 [4,-1,2,1] 的和最大,为 6...
https://leetcode-cn.com/problems/kth-largest-element-in-an-array 用普通的定长(k)数组作为容器,元素从小到大排列, 遍历列表中的元素,始终维持k个有效元素。 最后取第0个元素就是第k大的元素。 function H(/* int */cap) { this.repo = []; ...
11. Missing numberhttps://leetcode.com/problems/missing-number Given an arraynumscontainingndistinct numbers in the range[0, n], returnthe only number in the range that is missing from the array. sum(0:n) - sum(nums[0]:nums[n-1]) ...
38 awangdev/LintCode Java Solutions to problems on LintCode/LeetCode 4265 2 2022-04-04 39 NotFound9/interviewGuide 推荐刷题网站: https://www.lintcode.com/?utm_source=tf-github-NotFound2022 《大厂面试指北》——包括Java基础、JVM、数据库、mysql、redis、计算机网络、算法、数据结构、操作系统、设...
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/* 暴力解法时间复杂度:O(n³) 空间复杂度:O(1) 题解:https://leetcode-cn.com/problems/longest-palindromic-substring/solution/zui-chang-hui-wen-zi-chuan-by-leetcode-solution/ */ func longestPalindrome1(s string) string { length := len(s) if length < 2 { return s } maxLen, begin ...