I am trying to Implement a solutions to find k-th largest element in a given integer list with duplicates with O(N*log(N)) average time complexity in Big-O notation, where N is the number of elements in the list. As per my understanding Merge-sort has an average time complexity of O(N*log(N)) however in my below code I am actually using an extra for loop along with mergesort algorithm to delete
我试图实现一个解决方案，在给定的整数列表中找到第k个最大元素，其中O(N*log(N))平均时间复杂度为Big-O表示法，其中N是列表中元素的数量。 根据我的理解，合并排序的平均时间复杂度为O(N*log(N))但是在我的下面的代码中，我实际上使用了一个额外的for循环以及mergesort算法来删除重复，这肯定违反了我的find O(N*log(N))的第k个最大元素。 我如何通过在Big-O符号中实现我的任务O(N*log(N))平均时间复杂度来解决这个问题？ pu
Possible Duplicate: Plain English explanation of Big O I was recently asked about my knowledge of how to use Big O notation and I was stumped because I had never come across Big O before. I have read the Wikipedia page about Big O and looked at some of the questions posted in Stackoverflow but I just simply don't understand. My question: Can someone provide an explanation of Big O in t
可能重复： Big O的纯英文解释 最近我被问到关于如何使用Big O符号的知识，而且我之前因为从未遇到过Big O而感到难过。 我已经阅读了关于Big O的维基百科页面，并看了一些在Stackoverflow上发布的问题，但我只是不明白。 我的问题：有人能以最简单的形式提供Big O的解释，并提供如何在以下Java方法中使用它的示例： public int getScore(int[] dice) { int[][] dups; dups = possibleDups(dice); // Set ca
Say you have a linked list structure in Java. It's made up of Nodes: class Node { Node next; // some user data } and each Node points to the next node, except for the last Node, which has null for next. Say there is a possibility that the list can contain a loop - ie the final Node, instead of having a null, has a reference to one of the nodes in the list which came before it. W
假设你在Java中有一个链表结构。 它由节点组成： class Node { Node next; // some user data } 每个节点指向下一个节点，除了最后一个节点，下一个节点为空。 假设列表可能包含一个循环 - 也就是说，最终的节点（而不是一个空）会引用列表中的节点之前的节点。 什么是最好的写作方式 boolean hasLoop(Node first) 如果给定的节点是带有循环的列表中的第一个，则返回true否则返回false ？ 你怎么写，以便它需要
Java programs can be very memory hungry. For example, a Double object has 24 bytes: 8 bytes of data and 16 bytes of JVM-imposed overhead. In general, the objects that represent the primitive types are very expensive. The same happens for any collection in the Java Standard Library. There are even some counterintuitive facts such as a HashSet being more memory hungry than a HashMap , since a
Java程序可能非常渴望内存。 例如， Double对象有24个字节：8字节的数据和16字节的JVM开销。 一般来说，表示原始类型的对象非常昂贵。 对于Java标准库中的任何集合都会发生同样的情况。 因为HashSet里面包含一个HashMap （http://docs.oracle.com/javase/7/docs/api/java/util/HashSet。），所以甚至有一些违反直觉的事实，例如HashSet比HashMap更渴望内存。 HTML）。 在高性能设置中建模数据和对象的委派时，您能否提出
So say for example I'm going through an 'if' block and in this block, I am comparing the value of some number to a constant. Would it be more expensive like this: if( foo.getOb().getVal() == CONST_0 ) { .... } .... if( foo.getOb().getVal() == _CONST_N ) { .... } else .... OR: int x = foo.getOb().getVal(); if( x == CONST_0 ) { .... } .... if( x == _CONST_N ) {
所以说例如我正在经历一个'if'块，在这个块中，我将一些数字的值与一个常量进行比较。 它会像这样更昂贵： if( foo.getOb().getVal() == CONST_0 ) { .... } .... if( foo.getOb().getVal() == _CONST_N ) { .... } else .... 要么： int x = foo.getOb().getVal(); if( x == CONST_0 ) { .... } .... if( x == _CONST_N ) { .... } else .... 我知道这可能看起来像一个愚蠢的问题。
I have a question that when should we use Enum and when should we use a final constants? I know that it has been discussed at Enums and Constants. Which to use when? though it is C# question. My question is why Android use so many Constants rather than Enum? For example , Context In my opinion, if we use constants, there may be the risk that as below: if we define a LEVEL Constant that
我有一个问题，我们应该什么时候使用Enum，什么时候应该使用最终的常量？ 我知道它已经在Enums和Constants上讨论过了。 哪个使用时？ 尽管这是C＃的问题。 我的问题是为什么Android使用如此多的常量而不是Enum？ 例如，上下文 在我看来，如果我们使用常量，可能存在以下风险：如果我们定义一个LEVEL常量， public static final int LEVEL_LOW=1; public static final int LEVEL_MEDIUM=2; public static final int L
I know that I can perform divide by 2 using right shift. For simplicity, take a 4 bit number system -1 - 1111 -2 - 1110 -3 - 1101 -4 - 1100 -5 - 1011 -6 - 1010 -7 - 1001 -8 - 1000 7 - 0111 6 - 0110 5 - 0101 4 - 0100 3 - 0011 2 - 0010 1 - 0001 0 - 0000 If I try to perform 6 / 2 = 0110 >> 1 = 0011 = 3 -6/ 2 = 1010 >> 1 = 1101 = -3 Is valid for both +ve and -ve number How
我知道我可以用右移执行2分。 为了简单起见，需要一个4位数字系统 -1 - 1111 -2 - 1110 -3 - 1101 -4 - 1100 -5 - 1011 -6 - 1010 -7 - 1001 -8 - 1000 7 - 0111 6 - 0110 5 - 0101 4 - 0100 3 - 0011 2 - 0010 1 - 0001 0 - 0000 如果我尝试表演 6 / 2 = 0110 >> 1 = 0011 = 3 -6/ 2 = 1010 >> 1 = 1101 = -3 对于+ ve和-ve编号均有效 但是，当来到1 1 / 2 = 0001 >> 1 = 0000 = 0 -1/ 2 = 1
I encountered a question in interview to write a method to check for similar words irrespective of character cases. I answered it by using the difference of ASCII value for each pair of characters. But at home, when I went through the actual implementation of it in String.class, I get disturbed - Why is it implemented that way! I tried to draw a comparison between inbuilt and my custom metho
我在面试中遇到了一个问题，写出一种方法来检查类似的词语，而不考虑字符个案。 我通过使用每对字符的ASCII值的差异来回答它。 但在家里，当我在String.class中实际执行它时，我感到不安 - 为什么它以这种方式实现！ 我试图在内置和我的自定义方法之间进行比较， public class EqualsIgnoreCase { public static void main(String[] args) { String str1 = "Srimant @$ Sahu 959s"; String str2 = "sr
I have created the following function for checking the connection status: private void checkConnectionStatus() { HttpClient httpClient = new DefaultHttpClient(); try { String url = "http://xxx.xxx.xxx.xxx:8000/GaitLink/" + strSessionString + "/ConnectionStatus"; Log.d("phobos", "performing get " + url); HttpGet method = new HttpGet(new URI(url));
我创建了以下函数来检查连接状态： private void checkConnectionStatus() { HttpClient httpClient = new DefaultHttpClient(); try { String url = "http://xxx.xxx.xxx.xxx:8000/GaitLink/" + strSessionString + "/ConnectionStatus"; Log.d("phobos", "performing get " + url); HttpGet method = new HttpGet(new URI(url)); HttpResponse response = httpClient.exe
Here's my task which I want to solve with as little effort as possible (preferrably with QT & C++ or Java): I want to use webcam video input to detect if there's a (or more) crate(s) in front of the camera lens or not. The scene can change from "clear" to "there is a crate in front of the lens" and back while the cam feeds its video signal to my application. For
以下是我想尽可能少地努力解决的任务（最好使用QT＆C ++或Java）：我想使用网络摄像头视频输入来检测相机镜头前是否有（或更多）机箱或不。 场景可以从“清晰”变为“镜头前有一个箱子”，并且在凸轮将其视频信号传送给我的应用程序时返回。 对于原型测试/学习，我有2-3个“空”场景图像和2-3个带有一个或多个箱子的图像。 你知道如何解决这个问题吗？ 我发现了OpenCV，但是这个框架太笨重了吗？ 我是计算机视觉领域的新手。 这