smaller

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Have smaller server identifier, so dropping the connection: (2, 1)

有5个zk节点，有1个节点查看状态一直报错： ZooKeeper JMX enabled by default Using config: /home/hadoop/bigdata/zk/bin/../conf/zoo.cfg Client port found: 2181. Client address: localhost. Error contacting service. It is probably not running. 手工重启后，任然是上面的信

batch size 会影响训练的计算效率吗？

We also use a smaller mini-batch size of 256 without any noticeable performance degradation. This is in contrast to CURL and DrQ that both use a larger batch size of 512 to attain more stable training in the expense of computational efficiency. 　　　　　

python排序之快速排序

快速排序快速排序是比较常用的一种排序方式，通过递归的方法进行排序首先使用递归方式我们先要解决两个问题：1找到基准条件 2找到递归条件基线条件为数组为空或只包含一个元素。在这种情况下，只需原样返回数组——根本就不用排序那么这样我们就会把数组长度小于2的定位我们基

315. Count of Smaller Numbers After Self

数组中每个元素所在位置右边比它小的元素 int[] nums = { 5, 2, 6, 1 }; var rslt = CountSmaller(nums); foreach (var item in rslt) { Console.Write(item + " "); } private int[] CountSmaller(int[] nums) { if (nums == null || nums.Length == 0) { return new in

broker

A broker is a person or firm who arranges transactions between a buyer and a seller for a commission when the deal is executed. A broker who also acts as a seller or as a buyer becomes a principal party to the deal. Neither role should be confused with th

baggage

Baggage or luggage consists of bags, cases, and containers which hold a traveller's personal articles while the traveler is in transit. A modern traveller can be expected to have packages containing clothing, toiletries, small possessions, trip neces

恢复数组的方法数

整型数组arr长度为n(3 <= n <= 10^4)，最初每个数字是<=200的正数且满足如下条件： arr[0] <= arr[1] 2.arr[n-1] <= arr[n-2] arr[i] <= max(arr[i-1], arr[i+1]) 但是在arr有些数字丢失了，比如k位置的数字之前是正数，丢失之后k位置的数字为0。请你根据上述条件，计算可能有多少种不同

单调栈

当遇到的问题相当于，求解 prev smaller element ：初始数组从后往前，维护一个单调递增栈 next smaller element：初始数组从前往后，维护一个单调递增栈

快速排序算法总结

快速排序(Quick Sort)使用分治法策略。它的基本思想是：选择一个基准数，通过一趟排序将要排序的数据分割成独立的两部分；其中一部分的所有数据都比另外一部分的所有数据都要小。然后，再按此方法对这两部分数据分别进行快速排序，整个排序过程可以递归进行，以此达到整个数据变成有序序列。

315. Count of Smaller Numbers After Self 数组比自己小的元素个数，不能for要用bst

You are given an integer array nums and you have to return a new counts array. The counts array has the property where counts[i] is the number of smaller elements to the right of nums[i]. Example 1: Input: nums = [5,2,6,1] Output: [2,1,1,0] Explan

Alias采样算法

Alias采样过程分析学习来源： 1、时间复杂度O(1)的离散采样算法——Alias method 2、Alias采样算法程序实现 import numpy as np def alias_setup(probs): K = len(probs) q = np.zeros(K) J = np.zeros(K, dtype=np.int) smaller = [] larger = []

[LeetCode] 259. 3Sum Smaller tag: Two pointers

Given an array of n integers nums and an integer target, find the number of index triplets i, j, k with 0 <= i < j < k < n that satisfy the condition nums[i] + nums[j] + nums[k] < target. Example 1: Input: nums = [-2,0,1,3], target = 2 Ou

折半查找法 Java ＞＞＞算法

该方法仅适用于有序且唯一的数组（“数组”在Python中称为“列表”）如找到，则返回下标，否则，返回-1。 // 顺序折半查找下标 public static int BinarySearch(int key, int[] num_list) { int smaller = 0; int bigger = num_list.length - 1; while(smaller <= bigger){

用SAS进行泊松，零膨胀泊松和有限混合Poisson模型分析

原文链接：http://tecdat.cn/?p=6145 泊松模型 proc fmm data = tmp1 tech = trureg; model majordrg = age acadmos minordrg logspend / dist = truncpoisson; probmodel age acadmos minordrg logspend; /* Fit Statistics -2 Log Likelihood

【Rust日报】2020-07-31 Rust 1.45.1发布

Rust 1.45.1发布上次1.45稳定版发布后没多久有人在github爆出了一个似乎动摇到Rust根基的bug（可变借用失效），后来发现主要原因是之前一个重构和整理const propagator的PR引入的问题（具体见Rust Stable 1.45 中的神奇 Bug 解读），官方博客提到在发布release版之前进行的crater测试（crater测

函数的嵌套

1.什么是函数的嵌套"""　　互相嵌套的两个函数: 　　包裹在外层的叫做外函数,内层的就是内函数　　 1 def outer(): 2 # inner() 3 def inner(): 4 print("我是inner函数") (1)内部函数可以直接在函数外部调用么不行 inner() (2)调用外部函数后,内部函数

DistilBERT, a distilled version of BERT: smaller, faster, cheaper and lighter学习

1. 总结论文地址论文写得很简单，但是引用量好高啊

[LeetCode 315] Count of Smaller Numbers After Self

You are given an integer array nums and you have to return a new counts array. The counts array has the property where counts[i] is the number of smaller elements to the right of nums[i]. Example 1: Input: nums = [5,2,6,1] Output: [2,1,1,0] Explan

lower_bound 和 upper_bound

LeetCode 315. 计算右侧小于当前元素的个数 You are given an integer array nums and you have to return a new counts array. The counts array has the property where counts[i] is the number of smaller elements to the right of nums[i]. Example 1: Input: nums = [5,2

LeetCode #1365. How Many Numbers Are Smaller Than the Current Number

题目 1365. How Many Numbers Are Smaller Than the Current Number 解题方法设立一个temp数组存放nums中元素的有序序列，遍历temp中的每个元素i，累计已经遍历过的数的个数countnum，如果当前元素i不在dic中就加进去，值设置为已经遍历过的数的个数countnum，这个遍历结束之后遍历nums查

315. Count of Smaller Numbers After Self(二分或者算法导论中的归并求逆序数对)

You are given an integer array nums and you have to return a new counts array. The counts array has the property where counts[i] is the number of smaller elements to the right of nums[i]. Example 1: Input: nums = [5,2,6,1] Output: [2,1,1,0] Expl

最大公约数

计算两个数的最大公约数 1 def gcd(x, y): 2 '''该函数返回两个数的最大公约数''' 3 if not(x > 0 and y > 0): 4 print("Invalid numbers!") 5 # 用匿名函数简化构造取较小值的函数 6 smaller = lambda x,y: x if x < y else y 7

[LeetCode] 315. Count of Smaller Numbers After Self

You are given an integer array nums and you have to return a new counts array. The counts array has the property where counts[i] is the number of smaller elements to the right of nums[i]. Example 1: Input: nums = [5,2,6,1] Output: [2,1,1,0] Explana

[Java/Python]输出两数中的最小数 one-liner

Java import java.util.Scanner; public class compare { public static void main(String[] args) { Scanner scan = new Scanner(System.in); System.out.println("input your first number: "); int a = scan.nextInt();

CV第五课 Convolutional Neural Networks 下

1. 池化层 pooling 特征降维，减少过拟合　　1) pooling的目的是使得每一层smaller and more manageable ，这是为了有更少的参数数量，来表示输入的这张图　　2）池化操作只在平面上（actication map）操作，不在depth上操作，所以depth 不变　　3）它降低了采样率 2. max