Suppose you have a random list of people standing in a queue. Each person is described by a pair of integers (h, k), where h is the height of the person and k is the number of people in front of this person who have a height greater than or equal to …

lc 406 Queue Reconstruction by Height 406 Queue Reconstruction by Height Suppose you have a random list of people standing in a queue. Each person is described by a pair of integers(h, k), where h is the height of the person and k is the number of pe…

用python实现多级排序,可以像C语言那样写个my_cmp,然后在sort的时候赋给参数cmp即可 但实际上,python处理cmp 是很慢的,因为每次比较都会调用my_cmp:而使用key和reverse就快得多,因为只需要在排序开始时处理一次即可,因此在排序的时候能不用cmp就尽量不用 另外可以用operator函数中的itemgetter,和attrgetter实现基于key的(多级)排序: from operator import itemgetter, attrgetter sort…

Suppose you have a random list of people standing in a queue. Each person is described by a pair of integers(h, k), where h is the height of the person and k is the number of people in front of this person who have a height greater than or equal to h…

一开始backtrack,设计了很多剪枝情况,还是TLE了 ..后来用PQ做的. 其实上面DFS做到一半的时候意识到应该用PQ做,但是不确定会不会TLE,就继续了,然后果然TLE了.. PQ的做法和剪枝的判断标准是一样的,只不过PQ不是遍历,比兼职纯遍历要快. 想象一下排座位,h是身高,k是某个小朋友最多允许几个人挡住他(貌似不是很恰当). 对于单个小朋友来说,只有比他高才会挡住他,所以我们先按h的标准排最高的一批小朋友,这样一来后面批次的小朋友不会影响自己,因为后来比自己矮的不可能挡住自己.…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 日期 题目地址:https://leetcode.com/problems/queue-reconstruction-by-height/#/description 题目描述 Suppose you have a random list of people standing in a queue. Each person is described b…

https://leetcode.com/contest/6/problems/queue-reconstruction-by-height/ 分析:每个表示成(a,b)的形式,其实找第一个,就是b为0,a最小的那一个,然后一次更新小于等于a的节点的序号b,就是b--,然后重复上面的过程. 我刚开始是按a排序,然后再按b排序,使用set来做,发现我还要修改set元素的值,发现不能修改,(这里还是对数据结构掌握的不透彻),左后使用vector来做,不断的进行排序,来维护顺序.代码写的有点乱,感觉还…

假设有打乱顺序的一群人站成一个队列. 每个人由一个整数对(h, k)表示,其中h是这个人的身高,k是排在这个人前面且身高大于或等于h的人数. 编写一个算法来重建这个队列.注意:总人数少于1100人.示例输入:[[7,0], [4,4], [7,1], [5,0], [6,1], [5,2]]输出:[[5,0], [7,0], [5,2], [6,1], [4,4], [7,1]] 详见:https://leetcode.com/problems/queue-reconstruction-by-h…

原题 思路: 一开始完全没有思路..看了别人的思路才解出来. 先按照他们的高度从高到低(因为我后面用的从前往后遍历插入,当然也可以从低到高)排序,如果高度一样,那么按照k值从小到大排序. 排完序后我们可以注意到这样一个事实:如果我们按照先处理身最高的,那他们的k值就是他们所应该在的位置--因为已经没有比他们更高的了. 所以我们从高度从高到低按照k值的位置一直插入到答案中即可. class Solution { public: static bool cmp(pair<int, int> a,…

406. 根据身高重建队列 406. Queue Reconstruction by Height 题目描述 假设有打乱顺序的一群人站成一个队列.每个人由一个整数对 (h, k) 表示,其中 h 是这个人的身高,k 是排在这个人前面且身高大于或等于 h 的人数.编写一个算法来重建这个队列. 注意: 总人数少于1100人. 每日一算法2019/6/18Day 46LeetCode406. Queue Reconstruction by Height 示例: 输入: [[7,0], [4,4], […

Suppose you have a random list of people standing in a queue. Each person is described by a pair of integers(h, k), where h is the height of the person and k is the number of people in front of this person who have a height greater than or equal to h…

Suppose you have a random list of people standing in a queue. Each person is described by a pair of integers (h, k), where h is the height of the person and k is the number of people in front of this person who have a height greater than or equal to …

题目如下: Suppose you have a random list of people standing in a queue. Each person is described by a pair of integers (h, k), where h is the height of the person and k is the number of people in front of this person who have a height greater than or equ…

Suppose you have a random list of people standing in a queue. Each person is described by a pair of integers (h, k), where h is the height of the person and k is the number of people in front of this person who have a height greater than or equal to …

Description Suppose you have a random list of people standing in a queue. Each person is described by a pair of integers (h, k), where h is the height of the person and k is the number of people in front of this person who have a height greater than…

Level:   Medium 题目描述: Suppose you have a random list of people standing in a queue. Each person is described by a pair of integers (h, k), where h is the height of the person and k is the number of people in front of this person who have a height gre…

这题的关键点在于对数组的重排序方法,高度先由高到低排列不会影响第二个参数,因为list.add的方法在指定index后面插入,因此对于同高的人来说需要对第二个参数由低到高排,具体代码如下 public class Solution { public int[][] reconstructQueue(int[][] people) { Arrays.sort(people, new Comparator<int[]>() { public int compare(int[] o1, int[]…

终于将LeetCode的免费题刷完了,真是漫长的第一遍啊,估计很多题都忘的差不多了,这次开个题目汇总贴,并附上每道题目的解题连接,方便之后查阅吧~ 477 Total Hamming Distance 44.10% Meidum 475 Heaters  30.20% Easy 474 Ones and Zeroes  34.90% Meidum 473 Matchsticks to Square  31.80% Medium 472 Concatenated Words 29.20% Hard…

463. Island Perimeterhttps://leetcode.com/problems/island-perimeter/就是逐一遍历所有的cell,用分离的cell总的的边数减去重叠的边的数目即可.在查找重叠的边的数目的时候有一点小技巧,就是沿着其中两个方向就好,这种题目都有类似的规律,就是可以沿着上三角或者下三角形的方向来做.一刷一次ac,但是还没开始注意codestyle的问题,需要再刷一遍. class Solution { public: int islandPerime…

2.Add Two Numbers 原题链接https://leetcode.com/problems/add-two-numbers/ AC解: public ListNode addTwoNumbers(ListNode l1, ListNode l2) { int sum = 0; ListNode head = new ListNode(0); ListNode dummy = new ListNode(0); ListNode flag = dummy; dummy.next = he…

首先表达我对leetcode网站的感谢,与高校的OJ系统相比,leetcode上面的题目更贴近工作的需要,而且支持的语言广泛.对于一些比较困难的题目,可以从讨论区中学习别人的思路,这一点很方便. 经过一段时间的练习,我感觉自己的算法水平还是有很大的提升的.与学校开的算法导论的课程相比,我觉得实实在在的做题,比学习理论更符合我的认知方式.可能我属于那种脑子记不住,但是可以用“肌肉”去记忆的类型吧. 在做题的时候,我会尝试用感觉好玩的语言来写.如果把做题当成一种解谜游戏的话,那语言就是玩这个游戏选择…

突然很想刷刷题,LeetCode是一个不错的选择,忽略了输入输出,更好的突出了算法,省去了不少时间. dalao们发现了任何错误,或是代码无法通过,或是有更好的解法,或是有任何疑问和建议的话,可以在对应的随笔下面评论区留言,我会及时处理,在此谢过了. 过程或许会很漫长,也很痛苦,慢慢来吧. 编号 题名 过题率 难度 1 Two Sum 0.376 Easy 2 Add Two Numbers 0.285 Medium 3 Longest Substring Without Repeating C…

2017/11/14 Leetcode 日记 442. Find All Duplicates in an Array Given an array of integers, 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once. Find all the elements that appear twice in this array. Could you do it withou…

算法思想 二分查找 贪心思想 双指针 排序 快速选择 堆排序 桶排序 搜索 BFS DFS Backtracking 分治 动态规划 分割整数 矩阵路径 斐波那契数列 最长递增子序列 最长公共子系列 0-1 背包 数组区间 字符串编辑 其它问题 数学 素数 最大公约数 进制转换 阶乘 字符串加法减法 相遇问题 多数投票问题 其它 数据结构相关 栈和队列 哈希表 字符串 数组与矩阵 1-n 分布 有序矩阵 链表 树 递归 层次遍历 前中后序遍历 BST Trie 图 位运算 参考资料 算法思想 二…

贪心基础 贪心(Greedy)常用于解决最优问题,以期通过某种策略获得一系列局部最优解.从而求得整体最优解. 贪心从局部最优角度考虑,只适用于具备无后效性的问题,即某个状态以前的过程不影响以后的状态.紧接下来的状态仅与当前状态有关.和分治.动态规划一样,贪心是一种思路,不是解决某类问题的具体方法. 应用贪心的关键,是甄别问题是否具备无后效性.找到获得局部最优的策略.有的问题比较浅显,例如一道找零钱的题目 LeetCode 860. Lemonade Change: // 860. Lemonad…

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…

All LeetCode Questions List 题目汇总 Sorted by frequency of problems that appear in real interviews. Last updated: October 2, 2017Google (214)534 Design TinyURL388 Longest Absolute File Path683 K Empty Slots340 Longest Substring with At Most K Distinct C…

源代码地址:https://github.com/hopebo/hopelee 语言:C++ 301. Remove Invalid Parentheses Remove the minimum number of invalid parentheses in order to make the input string valid. Return all possible results. Note: The input string may contain letters other tha…

307. Range Sum Query - Mutable 思路:利用线段树,注意数据结构的设计以及建树过程利用线段树,注意数据结构的设计以及建树过程 public class NumArray { class segmentNode{ int start; int end; segmentNode left; segmentNode right; int sum; public segmentNode(int start,int end){ this.start = start; this.…

[LeetCode] Top 100 Liked Questions # Title Acceptance Difficulty 1 Two Sum 38.80% Easy 2 Add Two Numbers 29.10% Medium 3 Longest Substring Without Repeating Characters 25.00% Medium 4 Median of Two Sorted Arrays 23.70% Hard 5 Longest Palindromic Subs…