sharepoint online – How to display a list item and its linked items in the easiest way?

In SharePoint 2013 and earlier, you could have used a URL parameter to filter the Web Part of the Project List based on the client ID, but at the present time, Modern SharePoint does not have Web Parts or the functionality to do so.

However, you can use an integrated PowerApp. Use the client as a data source, use a drop-down menu to select the client, and then use the projects as another data source, filter based on the selected client ID, and view the results in a gallery.

java – Reverse linked list

I try to reverse the list without changing the content of the nodes. Ex: 1, 2, 3, 4 = 4, 3, 2, 1. I tried some things and I could not, could anyone help me?
The code is as follows:

//iniciamente a minha lista não aponta para lugar algum
public static Nodo inicio = null;

public static void print() {
    if (inicio == null) {
        System.out.println("Nada a exibir - a lista está vazia!");
    } else {
        Nodo aux = inicio;
        while (aux != null) {
            System.out.print(aux.getChave() + " ");
            aux = aux.getProx();
        }
    }
}

public void inserir(int e) {
    //criar um novo Nodo
    Nodo novo = new Nodo();
    novo.setChave(e);    //inserindo elemento
    novo.setProx(null); //depois dele não vem ninguém

    if (inicio == null) {
        inicio = novo; //aponto para o novo nodo
    } else {     //ja tem gente na lista ai precisa percorrer até chegar na null
        Nodo aux = inicio; //joga o aux no inicio
        while (aux.getProx() != null) {
            aux = aux.getProx();
        }
        aux.setProx(novo);
    }
}

public static void inverter() {
}

public boolean isEmpty() {
    return this.inicio == null;
}

public static void main(String() args) {
    Main m = new Main();
    m.inserir(1);
    m.inserir(2);
    m.inserir(3);
    m.print();
}

java – How to remove the last item from a linked list using position?

I want to delete the last element of a linked list by entering position n:

when I type 1,2,3,4,5 do and n = 2, I want it to be 1,2,3,5,5 but it gives 1,2,3,5.

import java.io.IOException;
import java.util.Scanner;

class No {
    int val;
    No prox;

    No(int x) {
        val = x;
    }
}

public class Main {
    public static No removeNElementoDoFim(No cabeca, int n) {

        if (cabeca == null) {
            return null;
        } else {
            No ultimoNo = cabeca.prox;
            No penultimoNo = cabeca.prox;
            for (int i = 0; ultimoNo.prox != null && i < n; i++) {
                penultimoNo = ultimoNo;
                ultimoNo = ultimoNo.prox;
            }
            penultimoNo.prox = null;
            return cabeca;
        }

    }

    public static int() stringToIntegerArray(String input) {
        input = input.trim();
        if (input.length() == 0) {
            return new int(0);
        }

        String() parts = input.split(",");
        int() output = new int(parts.length);
        for (int index = 0; index < parts.length; index++) {
            if (parts(index).trim().length() > 0) {
                String part = parts(index).trim();
                output(index) = Integer.parseInt(part);
            }
        }
        return output;
    }

    public static No stringToNo(String input) {

        int() nodeValues = stringToIntegerArray(input);

        No root = new No(0);
        No ptr = root;
        for (int item : nodeValues) {
            ptr.prox = new No(item);
            ptr = ptr.prox;
        }
        return root.prox;
    }

    public static String noToString(No no) {
        if (no == null) {
            return "()";
        }

        String result = "";
        while (no != null) {
            result += Integer.toString(no.val) + ", ";
            no = no.prox;
        }
        return "(" + result.substring(0, result.length() - 2) + ")";
    }

    public static void main(String() args) throws IOException {
        Scanner scann = new Scanner(System.in);
        String line;
        while (scann.hasNextLine()) {
            line = scann.nextLine();
            No cabeca = stringToNo(line);

            int n = Integer.valueOf(scann.nextLine());

            No ret = removeNElementoDoFim(cabeca, n);

            String out = noToString(ret);

            System.out.print(out);
        }

        scann.close();
    }
}

data structures – Parallel deletion and traversal in a linked list free of any lock

I am not sure of the security of deletion in a linked list. I've seen the problem posed by parallel insertion and deletion (resulting in loss of insertion) and proposed solutions to this problem. It's perfectly logical. However, I have not seen anywhere mentioning that an unfortunate crossing in parallel with a deletion can cause problems. For example, if we have the linked list:

A -----> B -----> C* ------> D

and a process (process 1) passes through it, is currently pointing to C (as indicated by *), however, before process 1 reads the next C pointer, another process (process 2) deletes C and frees its memory. Then, when the process 1 tries to read the next C pointer, it will read an uninitialized memory (which could be used to store an int for example), and could therefore be pointed at an invalid location and cause a segmentation error by dereferencing needle.

How to avoid this problem?

Linked SharePoint list that does not produce correct data in a REST request?

I am trying to use the Office 365 SharePoint REST API to extract the data stored in a list on SharePoint.

There are 2 columns in the list that look for another list.

I'm trying to query the list using:

.../site/_api/web/lists/getbytitle('Sample Lists')/items?$select=Email/Title&$expand=Email

The Email column of the list stores a search in a user table that then stores the e-mail. However, when I run the development in this way, I do not get a list of emails, but a list of names (the other search value from the original list).

construction of the function – automatically linked histogram regions?

Since you work with fairly dense histogram data, you can also play with the smooth kernel density estimator of your histogram:

SmoothHistogram[sample]

enter the description of the image here

dist = SmoothKernelDistribution[sample];
cddd = PDF[dist, x];
boundary1 = FindMinimum[{cddd, x > .995, x < 1}, {x, .995}] // Quiet
boundary2 = FindMinimum[{cddd, x > 1, x < 1.005}, {x, 1}] // Quiet

{23.8826, {x -> 0.995819}}

{18.3624, {x -> 1.00092}}

You can also play with the density estimator to find a more precise cut point:

SmoothHistogram[sample, .0002]

enter the description of the image here

dist2 = SmoothKernelDistribution[sample, .0002];
cddd2 = PDF[dist2, x];
boundary12 = FindMinimum[{cddd2, x > .995, x < 1}, {x, .995}] // Quiet
boundary12 = FindMinimum[{cddd2, x > 1, x < 1.005}, {x, 1}] // Quiet

{2.54235, {x -> 0.995272}}

{0.352377, {x -> 1.00102}}

If you know a good way to get FindMinimum return all the minimums in a given range that will work better than the parcel departure conditions

c # – Inserting elements from the linked list to the "right" location

I need my recorder middlewares (nodes of a linked list) in the right order so that they complement and work as you wish.

All known middleware can be inserted at any time but the first and the last. They will always be there. This means that I always start with two of them and that others will have to find their place somewhere in between.

API

I came up with the following algorithm. middlewhere is some current position. From there I navigate in the beginning, list them and Zip another collection shifted by one. Then using the order dictionary I check if insert is after or equal current and before or even next (If multiple instances of the same middleware are used, they are equal). When I find the post, I call InsertNext and the job is done.

public static Middleware InsertRelative(this Middleware middleware, Middleware insert, IDictionary order)
{
    if (middleware.Previous is null && middleware.Next is null)
    {
        throw new ArgumentException("There need to be at least two middlewares.");
    }
    var first = middleware.First();
    var zip = first.Enumerate(m => m.Next).Zip(first.Enumerate(m => m.Next).Skip(1), (current, next) => (current, next));

    foreach (var (current, next) in zip)
    {
        var canInsert =
            order(insert.GetType()) >= order(current.GetType()) &&
            order(insert.GetType()) <= order(next.GetType());
        if (canInsert)
        {
            return current.InsertNext(insert);
        }
    }

    return default; // This should not never be reached.
}

Demo

Here is a demo of work. I use three additional aids: First, Last and Enumerate at navigate chain. The other classes are null which represent the types of middlewares of the other question or news that I have since created.

void Main()
{   
    var middlewareOrder = new()
    {
        typeof(PropertySetter),
        typeof(Stopwatch),
        typeof(Attachment),
        typeof(Lambda),
        typeof(Scope),
        typeof(Serializer),
        typeof(Filter),
        typeof(Transaction),
        typeof(Echo),
    };

    var positions = middlewareOrder.Select((m, i) => (m, i)).ToDictionary(t => t.m, t => t.i);

    positions.Dump();

    // Default configuration.
    var middleware = new PropertySetter().InsertNext(new Echo());

    // Insert some new middlewares in an arbitrary order.
    middleware
        .InsertRelative(new Attachment(), positions)
        .InsertRelative(new Serializer(), positions)
        .InsertRelative(new Scope(), positions)
        .InsertRelative(new PropertySetter(), positions);

    // Show their order.
    middleware.First().Enumerate(m => m.Next).Select(m => m.GetType()).Dump();
}

public abstract class Middleware
{
    public Middleware Previous { get; private set; }

    public Middleware Next { get; private set; }

    public T InsertNext(T next) where T : Middleware
    {
        (next.Previous, next.Next, Next) = (this, Next, next);
        return next;
    }
}

public static class LoggerMiddlewareExtensions
{
    public static Middleware InsertRelative(this Middleware middleware, Middleware insert, IDictionary order)
    {
        if (middleware.Previous is null && middleware.Next is null)
        {
            throw new ArgumentException("There need to be at least two middlewares.");
        }
        var first = middleware.First();
        var zip = first.Enumerate(m => m.Next).Zip(first.Enumerate(m => m.Next).Skip(1), (current, next) => (current, next));

        foreach (var (current, next) in zip)
        {
            var canInsert =
                order(insert.GetType()) >= order(current.GetType()) &&
                order(insert.GetType()) <= order(next.GetType());
            if (canInsert)
            {
                return current.InsertNext(insert);
            }
        }

        return default;
    }

    public static Middleware First(this Middleware middleware)
    {
        return middleware.Enumerate(m => m.Previous).Last();
    }

    public static Middleware Last(this Middleware middleware)
    {
        return middleware.Enumerate(m => m.Next).Last();
    }

    public static IEnumerable Enumerate(this Middleware middleware, Func direction)
    {
        do
        {
            yield return middleware;
        } while (!((middleware = direction(middleware)) is null));
    }
}

public class PropertySetter : Middleware { }
public class Stopwatch : Middleware { }
public class Attachment : Middleware { }
public class Lambda : Middleware { }
public class Scope : Middleware { }
public class Serializer : Middleware { }
public class Filter : Middleware { }
public class Transaction : Middleware { }
public class Echo : Middleware { }

Questions

I do not think I'll have hundreds of middlewares for these few assistants to not affect the performance very satisfactorily, but maybe you can still think of something smarter?

java- how to implement a pop operation in a stack implemented using a linked list?

I currently use Java to complete a test module that involves implementing a stack using the linked list.

I have tried to implement push () operation but I have not got the correct result for pop () operation.
I have tried as follows:

public class LLStack implements LLStackQInterface{
SintNode head = new SintNode(0);
@Override
public void push(int num) {

head.nextNode = new SintNode(num);
head = head.nextNode;
}

@Override
public int pop() {

    SintNode t = head;
    while(t.nextNode!=null)
    {
        t=t.nextNode;

    }
    t = null;
    return 0;

}

I have to push the last element so that I navigate to the last element of the list using pointer but I did not get the result!

I receive this as a result:

Failed to enter: PUSH 1 PUSH 3 POP PUSH 4 PUSH 6
POP PUSH 8 PUSH 9 POP
The expected output is: 8-> 4-> 1-> NULL

The actual output generated by your code: 9-> 8-> 6-> 4-> 3-> 1-> NULL

I managed to reverse the list, but it is clear that my pop () does not work. What should I do?

Has anyone promoted his forums via paid advertisements? Google ads? Facebook ads? Linked to? | Promotion Forum

I'm trying to put a little bit of word on my marketing forum and I wondered if anyone already had paid promotion experience via the "usual" sites: Facebook, Google, linkedin. Experiments with that?

javascript – Invert a linked list instead

I solve programming problems to learn Javascript. The challenge I have chosen is:

Given the head of a linked list, reverse it on the spot.

Here is my implementation in JS

// Given the head of a singly linked list, reverse it in-place.
const ReverseLinkedList = {
    reverseLinkedList: (/** @type{ListNode} */ head) => {
        if (head === undefined) {
            return undefined;
        }

        if (head.next === undefined) {
            return head;
        }

        let temp = head;
        let tail = head.next.next;

        head = temp.next;
        head.next = temp;
        head.next.next = undefined;

        while (tail !== undefined) {
            let tempNext = tail.next;

            temp = head;
            head = tail;
            head.next = temp;

            tail = tempNext;
        }

        return head;
    },
    ListNode: function ListNode(/** @type{ListNode} */ next, val) {
        return {
            next: next,
            val: val
        }
    }
};

module.exports = ReverseLinkedList;

and my test class

const expect = require('chai').expect;
const reverseLinkedList = require('../../src/reverseLinkedList');

let head;
let node_01;
let node_02;
let reversed;

node_02 = reverseLinkedList.ListNode(undefined, 2);
node_01 = reverseLinkedList.ListNode(node_02, 1);
head = reverseLinkedList.ListNode(node_01, 0);

reversed = reverseLinkedList.reverseLinkedList(head);
expect(reversed.val).to.equal(2);
expect(reversed.next.val).to.equal(1);
expect(reversed.next.next.val).to.equal(0);
expect(reversed.next.next.next).to.equal(undefined);

node_01 = reverseLinkedList.ListNode(undefined, 1);
head = reverseLinkedList.ListNode(node_01, 0);
head.next = node_01;

reversed = reverseLinkedList.reverseLinkedList(head);
expect(reversed.val).to.equal(1);
expect(reversed.next.val).to.equal(0);
expect(reversed.next.next).to.equal(undefined);

head = reverseLinkedList.ListNode(undefined, 0);
reversed = reverseLinkedList.reverseLinkedList(head);
expect(reversed.val).to.equal(0);
expect(reversed.next).to.equal(undefined);

Being a beginner in Javascript, do you have any comments on style and best practices? All comments on the algorithm itself are also highly appreciated.

Thank you.