我很困惑解决这个时间复杂性问题.
T(n) = T(n-1)
我知道快速排序最糟糕的情况 T(n) = T(n-1) + T(1) + n
评估为(n-1) + (n-2) + (n-3) + ... + 1&此几何序列等于O(n^2)
然而.我看到stackoverflow上的答案T(n) = T(n-1) + c = O(n).
当这也等于时(n-1) + (n-2) + (n-3) + ... + 1,这怎么可能呢？O(n^2)
有人可以解释一下.
每当我运行同时包含 theLIMIT和ORDER BY关键字的查询时，根据LIMIT子句中指定的项目数，我会得到不一致的结果：
例如，以下查询：
SELECT "people"."name"
FROM "people"
WHERE "people"."type" = 2
ORDER BY score desc, number desc
LIMIT 50
结果给了我这个：
[John, Jack, Amy, Brad, Tom, ...]
（约翰和杰克的score和number属性值相同。事实上，艾米、布拉德和汤姆也是如此。）
现在，当我运行时：
SELECT "people"."name"
FROM "people"
WHERE "people"."type" = 2
ORDER BY score desc, number desc
LIMIT 5
我得到以下结果：
[Jack, John, Tom, Amy, Brad]
我有5个型号.Server,Platform,Game,RetentionReport,DataReport.而我正在尝试使用:dependent => :delete_all,但它不会工作.这是我的模特.
class Game < ActiveRecord::Base
attr_accessible :name
has_many :platforms, :dependent => :delete_all
end
class Platform < ActiveRecord::Base
attr_accessible :name, :game_id, :company_id
belongs_to :game
has_many :servers, :dependent => :delete_all
end
class Server < ActiveRecord::Base
attr_accessible :name, :region, :device_type, :platform_id, :platform_server_id
belongs_to :platform
has_many :gm_data_reports, :dependent => :delete_all
has_many :gm_retention_reports, :dependent => :delete_all
delegate :company_id, :to => :platform
validates :platform_server_id, :uniqueness => {:scope => :platform_id}
end
class DataReport …出于某种原因,我无法删除属于多对多关系的对象.我收到以下错误:
Exception in thread "main" org.hibernate.ObjectDeletedException: deleted object would be re-saved by cascade (remove deleted object from associations): [edu.cs157b.hibernate.AppointmentRequest#11]
at org.hibernate.internal.SessionImpl.forceFlush(SessionImpl.java:1232)
这是我的三个类,它们映射了多对多的关系.从本质上讲,Doctor有很多Patients通过AppointmentRequest,反之亦然.这是课程
医生
package edu.cs157b.hibernate;
import java.util.ArrayList;
import java.util.List;
import javax.persistence.*;
@Entity
@Table(name="DOCTOR_INFO")
@NamedQueries (
{
@NamedQuery(name = "Doctor.getAll", query = "from Doctor"),
@NamedQuery(name = "Doctor.findByName", query = "from Doctor where name = :name")
}
)
public class Doctor implements Person {
private int id;
private String name;
private Specialty specialty;
private List<AppointmentRequest> appointmentRequests …以下功能的时间复杂度是多少？
for(int i = 0; i < a.size; i++) {
for(int j = i; j < a.size; i++) {
//
}
}
我认为它不是大O n ^ 2因为我们没有迭代第二个for循环中的所有元素.我相信时间的复杂性是这样的:
n[ (n) + (n-1) + (n-2) + ... + (n-n) ]
但是,当我解决这个公式时,它就会出现
n^2 - n + n^2 - 2n + n^2 - 3n + ... + n^2 - n^2
这根本不正确.有人能告诉我究竟如何解决这个问题,以及我错在哪里.
嗨,我正在尝试更改select元素的CSS ,它在Firefox中工作正常,但它在谷歌浏览器中不起作用.当我试图使用Chrome开发者工具,我看到下面Styles/Matched CSS Rules的元素高度成功更改30px,但是当我去Computed Style了height显示为18px,但是当我点击箭头以获得更多的信息,它表明高度应30px
#new-job select - 30px
我是html/css和chromes开发人员工具的新手,但它似乎应该对我有用.这是我的代码
<div id="new-job">
<form accept-charset="UTF-8" action="/jobs" class="new_job" id="new_job" method="post"><div style="margin:0;padding:0;display:inline"><input name="utf8" type="hidden" value="?"><input name="authenticity_token" type="hidden" value="vfKmC+vnnIl/6icDIbDRfZghNR927qmEqTdB9I2OeMQ="></div>
<div class="field">
<label for="job_category">Category</label>
<select id="job_category" name="job[category]"><option value=""></option>
<option value="Bistro Card">Bistro Card</option>
<option value="Dongle">Dongle</option>
<option value="Return">Return</option>
<option value="Catering">Catering</option>
<option value="Other">Other</option></select><br><br>
</div>
<div class="field">
<label for="job_location">Location</label><br>
<input id="job_location" name="job[location]" size="30" type="text">
</div>
<div class="field">
<label for="job_description">Description</label><br>
<textarea cols="40" id="job_description" name="job[description]" rows="20"></textarea>
</div>
<div …我有一个由错误的控制器操作处理的URL.在我的路线文件中,我有以下内容
post '/tweets' => 'tweets#create'
每当我发布到这个网址时,它都会被处理tweets#index.
Started POST "/tweets" for 127.0.0.1 at 2014-03-27 20:04:25 -0700
Processing by TweetsController#index as HTML
Parameters: {"utf8"=>"?", "authenticity_token"=>"CCN9rDQiSXkVK0sF/BGU19fIufJTpho/ocySPmu7Lsc=", "tweet"=>{"title"=>"sdfdsafdsa", "description"=>"sdfdsasdadsa", "user_id"=>"1"}, "commit"=>"Create Tweet"}
Tweet Load (0.4ms) SELECT `tweets`.* FROM `tweets`
Rendered tweets/index.html.erb within layouts/application (0.1ms)
Completed 200 OK in 42ms (Views: 40.6ms | ActiveRecord: 0.4ms)
我试过重新启动我的rails服务器但它仍然无法正常工作.
这是结果 rake routes
tweeter: rake routes
new_user_session GET /users/sign_in(.:format) devise/sessions#new
user_session POST /users/sign_in(.:format) devise/sessions#create
destroy_user_session DELETE /users/sign_out(.:format) devise/sessions#destroy
user_password POST /users/password(.:format) devise/passwords#create
new_user_password GET /users/password/new(.:format) devise/passwords#new
edit_user_password …我正在实施一个小型社交网站,我正在尝试实施通知.
通知具有以下要求
以下是我想用MySQL实现它的方法:
通知表:type_of_event,event_id,message
Notification_read表:user_id,notification_id,read
Notification表中Notification_read表格将填写新创建的通知,其中包含新通知的ID和应接收通知的每个用户的ID(通知创建者的关注者)read这个解决方案对我来说似乎效率很低,因为每次发生通知时,通知都会notification_read多次写入表中(取决于用户拥有多少关注者)
有人能告诉我是否有更好的解决方案来解决这个问题
我正在尝试创建一个侦听套接字连接的进程.当我在main()函数中绑定,监听和等待接受时,它似乎有效.但是当我尝试创建一个新线程并在新线程上绑定,监听和接受时,它就失败了.这是我的代码.
void request_handler(int clientSock) {
FILE *requestedFile = NULL;
long fileSize = 0;
struct stat st;
long bytesRead;
char buffer[1024];
requestedFile = fopen("/PATH/book.txt", "rb");
while(!feof(requestedFile)) {
bytesRead = fread(buffer, 1, sizeof(buffer), requestedFile);
send(clientSock, buffer, bytesRead, 0);
}
}
void listener() {
int server_sock_desc;
struct sockaddr_in name;
int client_sock_desc;
struct sockaddr_in client_name;
socklen_t addr_size;
pthread_t handler_thread;
printf("waiting");
//connection setup
server_sock_desc = socket(PF_INET, SOCK_STREAM, 0);
if(server_sock_desc != -1) {
memset(&name, 0, sizeof(name));
name.sin_family = AF_INET;
name.sin_port = htons(5000);
name.sin_addr.s_addr = htonl(INADDR_ANY);
int …