LeetCode 886. Possible Bipartition】的更多相关文章

原题链接在这里:https://leetcode.com/problems/possible-bipartition/ 题目: Given a set of N people (numbered 1, 2, ..., N), we would like to split everyone into two groups of any size. Each person may dislike some other people, and they should not go into the s…
[LeetCode]886. Possible Bipartition 解题报告(Python) 作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 题目地址:https://leetcode.com/problems/possible-bipartition/description/ 题目描述: Given a set of N people (numbered 1, 2, ..., N), we would like to spli…
题目链接 886. 可能的二分法 题意: 给定一组 N 人(编号为 1, 2, ..., N), 我们想把每个人分进任意大小的两组. 每个人都可能不喜欢其他人,那么他们不应该属于同一组. 形式上,如果 dislikes[i] = [a, b],表示不允许将编号为 a 和 b 的人归入同一组. 当可以用这种方法将所有人分进两组时,返回 true:否则返回 false. 思路: 法一:DFS+染色 首先初始化所有节点的颜色为0,然后给没有染色的节点染上1,给这个节点的相邻节点染色上-1,以此类推,判…
Given a set of N people (numbered 1, 2, ..., N), we would like to split everyone into two groups of any size. Each person may dislike some other people, and they should not go into the same group. Formally, if dislikes[i] = [a, b], it means it is not…
All LeetCode Questions List(Part of Answers, still updating) 题目汇总及部分答案(持续更新中) Leetcode problems classified by company 题目按公司分类(Last updated: October 2, 2017) .   Top Interview Questions # Title Difficulty Acceptance 1 Two Sum Medium 17.70% 2 Add Two N…
DFS基础 深度优先搜索(Depth First Search)是一种搜索思路,相比广度优先搜索(BFS),DFS对每一个分枝路径深入到不能再深入为止,其应用于树/图的遍历.嵌套关系处理.回溯等,可以用递归.堆栈(stack)实现DFS过程. 关于广度优先搜索(BFS)详见:算法与数据结构基础 - 广度优先搜索(BFS) 关于递归(Recursion)详见:算法与数据结构基础 - 递归(Recursion) 树的遍历 DFS常用于二叉树的遍历,关于二叉树详见: 算法与数据结构基础 - 二叉查找树…
Given a set of N people (numbered 1, 2, ..., N), we would like to split everyone into two groups of any size. Each person may dislike some other people, and they should not go into the same group. Formally, if dislikes[i] = [a, b], it means it is not…
给定一组 N 人(编号为 1, 2, ..., N), 我们想把每个人分进任意大小的两组. 每个人都可能不喜欢其他人,那么他们不应该属于同一组. 形式上,如果 dislikes[i] = [a, b],表示不允许将编号为 a 和 b 的人归入同一组. 当可以用这种方法将每个人分进两组时,返回 true:否则返回 false. 示例 1: 输入:N = 4, dislikes = [[1,2],[1,3],[2,4]] 输出:true 解释:group1 [1,4], group2 [2,3] 示…
请点击页面左上角 -> Fork me on Github 或直接访问本项目Github地址:LeetCode Solution by Swift    说明:题目中含有$符号则为付费题目. 如:[Swift]LeetCode156.二叉树的上下颠倒 $ Binary Tree Upside Down 请下拉滚动条查看最新 Weekly Contest!!! Swift LeetCode 目录 | Catalog 序        号 题名Title 难度     Difficulty  两数之…
Given an undirected graph, return true if and only if it is bipartite. Recall that a graph is bipartite if we can split it's set of nodes into two independent subsets A and B such that every edge in the graph has one node in A and another node in B.…
题外话:LeetCode上一个测试用例总是通不过(我在文章末贴出通不过的测试用例),给的原因是超出运行时间,我拿那个测试用例试了下2.037ms运行完.我自己强行给加了这句: && max == && heightSize == ) ; 能力不行,不要脸来凑. 题目: 给定 n 个非负整数表示每个宽度为 1 的柱子的高度图,计算按此排列的柱子,下雨之后能接多少雨水. 上面是由数组 [0,1,0,2,1,0,1,3,2,1,2,1] 表示的高度图,在这种情况下,可以接 6 个…
In a group of N people (labelled 0, 1, 2, ..., N-1), each person has different amounts of money, and different levels of quietness. For convenience, we'll call the person with label x, simply "person x". We'll say that richer[i] = [x, y] if pers…
Given an undirected graph, return true if and only if it is bipartite. Recall that a graph is bipartite if we can split it's set of nodes into two independent subsets A and B such that every edge in the graph has one node in A and another node in B.…
原题链接在这里:https://leetcode.com/problems/is-graph-bipartite/ 题目: Given an undirected graph, return true if and only if it is bipartite. Recall that a graph is bipartite if we can split it's set of nodes into two independent subsets A and B such that eve…
[98]Validate Binary Search Tree [99]Recover Binary Search Tree [100]Same Tree [101]Symmetric Tree [104]Maximum Depth of Binary Tree [105]Construct Binary Tree from Preorder and Inorder Traversal [106]Construct Binary Tree from Inorder and Postorder T…
作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 公众号:负雪明烛 本文关键词:刷题顺序,刷题路径,好题,top100,怎么刷题,Leetcode, 力扣,Python, C++, Java 大家好,相信很多朋友在刷题时,看到 LeetCode 的 2000 多道题目,有点手足无措,非常需要一个刷题的顺序和清单. 我整理了在 LeetCode(中文版)上点赞数前 100 的题目,这些的题目基本都在千赞以上,全部都是好题. 举个例子,1…
作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 把自己刷过的所有题目做一个整理,并且用简洁的语言概括了一下思路,汇总成了一个表格. 题目的排列顺序是按照先Easy再Medium再Hard排列的,暂时还没有把题目全部整理完成.后序我会把刷过的所有的题目都整理到这个文档里. 题目 难度 解法 题目地址 566. Reshape the Matrix Easy 变长数组,求余法,维护行列计算在新的数组中的位置 https://blog.c…
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