implementation

Constructing a Binary tree in Java [closed]

谁说我不能喝 提交于 2019-11-30 11:35:20
I am constructing a binary tree. Let me know if this is a right way to do it. If not please tell me how to?? I could not find a proper link where constructing a general binary tree has been coded. Everywhere BST is coded. 3 / \ 1 4 / \ 2 5 This is the binary tree which i want to make.I should be able to do all the tree traversals.Simple stuff. public class Binarytreenode { public Binarytreenode left; public Binarytreenode right; public int data; public Binarytreenode(int data) { this.data=data; } public void printNode() { System.out.println(data); } public static void main(String ar[]) {

How to maximize the sum?

六眼飞鱼酱① 提交于 2019-11-30 10:40:08
We are given a sorted array. Let the initial value of pass be zero. We can do the following operation any number of times: Select any k numbers at a time. Add them all up. Add this sum to pass If a number, say x , is being selected for the first time from the array, then it is considered as x only. When it is selected for the second time , then it is considered as -x , and for the third time, again as x , and so on... For example, let the array be [-14, 10, 6, -6, -10, -10, -14] and k = 4 , and we'll only do the operation once. We select these 4 numbers: {14, 10, 6, -6} . Adding them up, we

Location Listener in Background Service Android

感情迁移 提交于 2019-11-30 09:42:40
Which is the better approach, directly implementing LocationListener like this public class BackgroundService extends Service implements LocationListener {} or normally declared the LocationListener inside the class? LocationListener locationListener = new LocationListener() {}; In the second piece of code you have to call the attribute locationListener prior to calling the methods of the interface. In the first piece of code you can access the interface methods directly. So if you know that every method call costs cpu time then to implement it directly in the class rather than putting it as

Implementing Barabasi-Albert Method for Creating Scale-Free Networks

半腔热情 提交于 2019-11-30 07:27:47
I'm trying to implement a very simple preferential attachment algorithm for creating scale-free networks. These have degree distributions that follow a power-law, i.e. P(k) ~ k^-g, where g is the exponent. The algorithm below should produce degree distributions with the exponent equal 3 +/- 0.1, my implementation does not the exponents are closer to 2.5 +/- 0.1. I'm clearly not understanding something somewhere and continue to get it wrong. I'm sorry if this is in the wrong place, I couldn't decide whether it should be in stackoverflow or maths.stackexchange.com. The Algorithm: Input: Number

How do I fix “wrong number of type arguments” while trying to implement a method?

你。 提交于 2019-11-30 06:04:14
问题 I'm trying to implement a method: struct Point<T> { x: T, y: T, } struct Line<T> { start: Point<T>, end: Point<T>, } impl Line { fn length(&self) -> f64 { let dx: f64 = self.start.x - self.end.x; let dy: f64 = self.start.y - self.end.y; (dx * dx + dy * dy).sqrt() } } fn main() { let point_start: Point<f64> = Point { x: 1.4, y: 1.24 }; let point_end: Point<f64> = Point { x: 20.4, y: 30.64 }; let line_a: Line<f64> = Line { start: point_start, end: point_end, }; println!("length of line_a = {}",

Implementation of Friend concept in Java [duplicate]

时间秒杀一切 提交于 2019-11-30 02:26:09
This question already has an answer here: Is there a way to simulate the C++ 'friend' concept in Java? 19 answers How does one implement the friend concept in Java (like C++)? Alastor Moody Java does not have the friend keyword from C++. There is, however, a way to emulate that; a way that actually gives a lot more precise control. Suppose that you have classes A and B. B needs access to some private method or field in A. public class A { private int privateInt = 31415; public class SomePrivateMethods { public int getSomethingPrivate() { return privateInt; } private SomePrivateMethods() { } //

Fastest implementation for All-pairs shortest paths problem?

末鹿安然 提交于 2019-11-29 19:42:21
问题 I have a weighted graph 30k nodes 160k edges, no negative weights. I would like to compute all the shortest paths from all the nodes to the others. I think I cannot assume any particular heuristics to simplify the problem. I tried to use this Dijkstra C implementation http://compprog.wordpress.com/2007/12/01/one-source-shortest-path-dijkstras-algorithm/, that is for a single shortest path problem, calling the function dijkstras() for all my 30 nodes. As you can imagine, it takes ages. At the

Where to find Python implementation of Chaikin's corner cutting algorithm?

旧巷老猫 提交于 2019-11-29 15:20:27
问题 I am looking for Chaikin's corner cutting algorithm (link1, link2) implemented in Python 2.7.X but can't find it. Maybe someone have it and able share the code? 回答1: Mr. Che answer will work, but here is a much shorter version that is slightly more efficient. import numpy as np def chaikins_corner_cutting(coords, refinements=5): coords = np.array(coords) for _ in range(refinements): L = coords.repeat(2, axis=0) R = np.empty_like(L) R[0] = L[0] R[2::2] = L[1:-1:2] R[1:-1:2] = L[2::2] R[-1] = L

Location Listener in Background Service Android

给你一囗甜甜゛ 提交于 2019-11-29 14:54:16
问题 Which is the better approach, directly implementing LocationListener like this public class BackgroundService extends Service implements LocationListener {} or normally declared the LocationListener inside the class? LocationListener locationListener = new LocationListener() {}; 回答1: In the second piece of code you have to call the attribute locationListener prior to calling the methods of the interface. In the first piece of code you can access the interface methods directly. So if you know

What are clever ways to output a list of n items with (n-1) separators inbetween?

爷,独闯天下 提交于 2019-11-29 14:24:38
问题 Let's say that we have an array with n elements (n > 0). We would like to output a list of those elements, with a separator between them. A common approach to this problem is: foreach item ( output item output separator ) trim last separator But it seems a bit messy to have to do that. Another approach would be: check that there is at least one element loop ( output element next element, or break if no more elements output separator ) But I am not sure that it will always work. Do you see