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240. Search a 2D Matrix II
Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties: Integers in each row are sorted i ...
分类:其他好文   时间:2020-11-25 12:59:50    阅读次数:14
Power of Two Choices 负载均衡
NGINX and the "Power of Two Choices" Load-Balancing Algorithm - NGINX https://www.nginx.com/blog/nginx-power-of-two-choices-load-balancing-algorithm/ ...
分类:其他好文   时间:2020-11-23 11:48:43    阅读次数:4
4. 寻找两个正序数组的中位数 Median of Two Sorted Arrays
Given two sorted arrays nums1 and nums2 of size m and n respectively, return the median of the two sorted arrays. The overall run time complexity shou ...
分类:编程语言   时间:2020-11-21 12:37:58    阅读次数:15
POJ 3368 Frequent values (ST表)
##题面 You are given a sequence of n integers a1 , a2 , ... , an in non-decreasing order. In addition to that, you are given several queries consisting ...
分类:其他好文   时间:2020-11-21 12:28:36    阅读次数:7
leetcode 116. Populating Next Right Pointers in Each Node
You are given a perfect binary tree where all leaves are on the same level, and every parent has two children. The binary tree has the following defin ...
分类:其他好文   时间:2020-11-20 12:00:39    阅读次数:9
29.shell脚本正整数加减乘除计算
思路:函数先计算数值加减乘除,case赋值输出对应的加减乘除#!/bin/bashadd(){add=$[$1+$2]echo"outcome:$1+$2=$add"}minus(){min=$[$1-$2]echo"outcome:$1-$2=$min"}multiply(){mult=$[$1*$2]echo"outcome:$1x$2=$mult"}divide(){div=$[$1/$2]
分类:系统相关   时间:2020-11-20 11:18:50    阅读次数:15
leetcode 1283. Find the Smallest Divisor Given a Threshold
Given an array of integers nums and an integer threshold, we will choose a positive integer divisor and divide all the array by it and sum the result ...
分类:其他好文   时间:2020-11-13 13:21:44    阅读次数:33
JavaSE---Concurrency
1、Processes and Threads In concurrent programming, there are two basic units of execution: processes and threads.(并发程序中,有2个基本的执行单位:进程、线程) 1.1、Processe ...
分类:编程语言   时间:2020-11-12 13:23:54    阅读次数:6
回调函数的理解
先看两个例子,比较不同: 例一(有回调函数) #include <iostream> using namespace std; void foo_one() { cout<<"foo_one"<<endl; } void foo_two() { cout<<"foo_two"<<endl; } ty ...
分类:其他好文   时间:2020-11-11 16:29:50    阅读次数:8
20.11.8 leetcode 1,122
#1:在数组中找两个数,使其和为target 链接:https://leetcode-cn.com/problems/two-sum/ 暴力就是n^2,可以用map来做,这里介绍一下map.count(tmp)函数,是搜索key的值,map里存在值为tmp的key就返回1,否则是0 #122:给定一 ...
分类:其他好文   时间:2020-11-08 17:59:16    阅读次数:25
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