st表是一种基于DP(动态规划)思想的算法,也算是一种数据结构吧。 st表可以静态维护区间的最值,需要用的时间来预处理,后可以O(1)查询。 我们定义dp[i][j]表示 表示区间的最值,在这道题里我们认为是最大值(维护最小值同理)。看一下能不能开下,大概是maxn * 20的大小,可以开下。 通过定义不难发现,dp...
转载:来自大佬https://leetcode-cn.com/problems/stone-game/solution/jie-jue-bo-yi-wen-ti-de-dong-tai-gui-hua-tong-yong/ 该问题主要是如何设计dp数组,其次是动态转移方程 1.dp数组设计 状态有三种:开始堆的索引i,结束堆的索引j,以及当前轮到的人 dp[i][j].fir 表示,对于 piles[i...j] 这部分...
This repository contains the solutions and explanations to the algorithm problems on LeetCode. Only medium or above are included. All are written in C++/Python and implemented by myself. The problems attempted multiple times are labelled with hyperlinks. Topics algorithms leetcode cpp Resources Re...
LeetCode R.I.P. to my old Leetcode repository, where there were 5.7k+ stars and 2.2k+ forks (ever the top 3 in the field). Since free questions may be even mistakenly taken down by some companies, only solutions will be post on now. There are new LeetCode questions every week. I...
参考题解https://leetcode-cn.com/problems/find-all-good-strings/solution/shu-wei-dp-kmp-by-qodjf/ 国服现在好多大神…题解写的真的很棒~ cal 计算小于等于任意字符串 s 的数目,计算两次求差就可了。其中 s1 没有被计算,所以单独算一次即可。
882. Maximal Square Medium 这个题使用dp,某位置的最大正方形边长等于左边、上边、左上边三个位置最小的边长+1 /fuxuemingzhu/article/details/82992233 883. Jump Game Medium 贪心算法,保存每一步都选择跳的最远的地方,如果这个位置不能到达直接返回false. /fuxuemingzhu/article/details/83504437 ...
j 可通过和任意数字异或得到,newDp[j] = 前一 dp 中最小的改变次数 + size[i]完成k 个组的动态规划后,dp[0]就是所求的解。 详细思路可参考:https://leetcode-cn.com/problems/make-the-xor-of-all-segments-equal-to-zero/solution/an-yu-shu-fen-zu-jin-xing-dong-tai-gui-h-7z0g/(...
1557 Minimum Number of Vertices to Reach All Nodes https://leetcode.com/problems/minimum-number-of-vertices-to-reach-all-nodes/ Given a directed acyclic graph, with n vertices numbered from 0 to n-1, ...1376. Time Needed to Inform All Employees 这道题很神奇,有些case输入,concele出不来...
关于上述代码的解析可参考力扣题解:https://leetcode-cn.com/problems/edit-distance/solution/bian-ji-ju-chi-by-leetcode-solution/ 遍历每个被查找的名称,计算它与数据库所有客户名称的编辑距离,并取编辑距离最小的客户名称: result = [] for name in find.name.values: ...
绿色即为后缀 <题目来自于力扣https://leetcode-cn.com/problems/all-paths-from-source-to-target人脸...