Question

我正在尝试将两个链接列表连接在一起，其中第二个列表将紧接在第一个列表的尾部之后。在我的append方法中，我想获得要连接在一起的两个列表，然后将最后一个列表连接到末尾。我在将当前位置分配到第二个列表的开头时遇到麻烦。关于下一步是什么建议？

public class Link {

public long dData;                 // data item
public Link next;                  // next link in list
// -------------------------------------------------------------

public Link(long d) // constructor
{
    dData = d;
}
// -------------------------------------------------------------

public void displayLink() // display this link
{
    System.out.print(dData + " ");
}
// -------------------------------------------------------------
}  // end class Link

public class FirstLastList {

private Link first;               // ref to first link
private Link last;                // ref to last link
// -------------------------------------------------------------

public FirstLastList() // constructor
{
    first = null;                  // no links on list yet
    last = null;
}
// -------------------------------------------------------------

public boolean isEmpty() // true if no links
{
    return first == null;
}
// -------------------------------------------------------------

public void insertFirst(long dd) // insert at front of list
{
    Link newLink = new Link(dd);   // make new link

    if (isEmpty()) // if empty list,
    {
        last = newLink;             // newLink <-- last
    }
    newLink.next = first;          // newLink --> old first
    first = newLink;               // first --> newLink
}
// -------------------------------------------------------------

public void insertLast(long dd) // insert at end of list
{
    Link newLink = new Link(dd);   // make new link
    if (isEmpty()) // if empty list,
    {
        first = newLink;            // first --> newLink
    } else {
        last.next = newLink;        // old last --> newLink
    }
    last = newLink;                // newLink <-- last
}
// -------------------------------------------------------------

public long deleteFirst() // delete first link
{                              // (assumes non-empty list)
    long temp = first.dData;
    if (first.next == null) // if only one item
    {
        last = null;                // null <-- last
    }
    first = first.next;            // first --> old next
    return temp;
}
// -------------------------------------------------------------

public void displayList() {
    System.out.print("List (first-->last): ");
    Link current = first;          // start at beginning
    while (current != null) // until end of list,
    {
        current.displayLink();      // print data
        current = current.next;     // move to next link
    }
    System.out.println("");
}
// -------------------------------------------------------------

public void append(FirstLastList list1, FirstLastList list2) {
    Link current = first;
    while (list1 != null) {
        current = current.next;
    }
    current.next = list2.first; 
}
}  // end class FirstLastList

public class FirstLastApp {

public static void main(String[] args) {                //make a new list
    FirstLastList theList = new FirstLastList();

    theList.insertFirst(22);       // insert at front
    theList.insertFirst(44);
    theList.insertFirst(66);

    theList.insertLast(11);        // insert at rear
    theList.insertLast(33);
    theList.insertLast(55);

    theList.displayList();         // display the list

    theList.deleteFirst();         // delete first two items
    theList.deleteFirst();

    theList.displayList();         // display again
    FirstLastList theList2 = new FirstLastList();

    theList.insertFirst(22);       // insert at front
    theList.insertFirst(44);
    theList.insertFirst(66);

    theList.insertLast(11);        // insert at rear
    theList.insertLast(33);
    theList.insertLast(55);

    theList.displayList();         // display the list

    theList.deleteFirst();         // delete first two items
    theList.deleteFirst();

    theList.displayList();         // display again

    append(theList, theList2);
    theList.displayList();         // display again


}  // end main()
}  // end class FirstLastApp

Answer 1

在

while (list1 != null) {
    current = current.next;
}

list1 不变，您将取消对NULL指针的引用

奇怪的是，当操作不是静态的并且不返回结果时，您在参数中获得了两个列表。

对于我来说，如果不是静态操作，该操作必须接收一个列表并将其追加到当前（此）列表的末尾，在参数中迭代列表并使用 insertLast < / p>

您还可以通过值接收参数，最好使用引用不复制任何内容

Answer 2

在append（）方法中：

---
title: "MWE Plotly"
output: html_document
---

```{r setup, include=FALSE}
knitr::opts_chunk$set(echo = TRUE)
#Needed packages
library(plotly)
library(ggcorrplot)
library(dplyr)
library(purrr)
#For reproducability
set.seed(1)

#Generate some numbers
numbers <- tibble(
  a = runif(10),
  b = runif(10),
  c = runif(10)
)

#Function for making the plot
cor_plotter_ex <- function(data){
  cor_plot <- ggcorrplot(corr = data)
  return(cor_plot)
}
```

```{r plot_plotly, out.width = "33%", hold = T, plotly = T}
#Some data
tib <- tibble(
  Id = letters[1:3],
  data = list(numbers)
) %>% 
  mutate(
    cor = map(data, ~cor(.x)),
    cor_plot = map(cor, ~cor_plotter_ex(.x))
  )

#Plot them with walk does not work
walk(tib$cor_plot,
     ~ggplotly(.x))
#As it only generates the first plot
#and it is not intractable
#Plot them with lapply and tagList does work
htmltools::tagList(lapply(tib$cor_plot,
     function(.x) {ggplotly(.x)}))
#Iff one renders the document with 
#rmarkdown::render('mwe_plotly.Rmd', output_file='report.html')
#Plotting one of them works
ggplotly(tib$cor_plot[[1]])
```

```{r regular_plot, out.width = "33%", hold = T}
#Just printing the plots clearly works
walk(tib$cor_plot,
     ~print(.x))
```

当前节点到达最后一个节点时，其Link current = first; while(current.next != null) { current = current.next; } current.next = list2.first;将为null。那就是您加入第二个列表的时候。

如何将链接列表附加到另一个列表的末尾？

2 个答案: