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…

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[]…

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…

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], […

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…

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 …

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…

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…

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,…