[LeetCode] Loud and Rich 聒噪与富有】的更多相关文章

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…
[LeetCode]851. Loud and Rich 解题报告(Python) 标签(空格分隔): LeetCode 作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 题目地址:https://leetcode.com/problems/loud-and-rich/description/ 题目描述: In a group of N people (labelled 0, 1, 2, -, N-1), each person ha…
851. Loud and Rich 题目链接:https://leetcode.com/problems/loud-and-rich/description/ 思路:有向图DFS,记录最小的quiet值 注意点:可优化,记忆性搜索,每次搜到已经记录过的值时可直接比较不需要进一步搜下去. 1 void DFS(vector<int> &visited, vector<vector<int> >& G, int x,vector<int>&am…
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…
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…
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…
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…
[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…
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…
iOS 基本内存管理-多对象内存管理(2)中可以看到涉及到对象的引用都要手动管理内存:每个对象都需要写如下代码 // 1.对要传入的"新车"对象car和目前Person类对象所拥有的"旧车"_car进行判读- (void)setCar:(Car *)car { if (_car != car ) { [_car release]; // 释放旧车 _car = [car retain]; // 新车引用计数加一 } } // 2.Person类在回收的时候也必须将它…