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在Java中如何在排序列表中插入元素_Java_List_Class_Sorting_Insert - Fatal编程技术网

在Java中如何在排序列表中插入元素

在Java中如何在排序列表中插入元素,java,list,class,sorting,insert,Java,List,Class,Sorting,Insert,我必须用java创建自己的列表。要插入的元素(Person元素)必须进行排序(从低到高)。 因此,我创建了一个方法compareTo来检查插入人员的姓名。 问题在于插入方法。我写了这段代码: public boolean insert(Person p){ PersonListElement element = new PersonListElement(p); boolean inserted = false; if(containsElement(p) == true

我必须用java创建自己的列表。要插入的元素(Person元素)必须进行排序(从低到高)。 因此,我创建了一个方法compareTo来检查插入人员的姓名。 问题在于插入方法。我写了这段代码:

public boolean insert(Person p){
    PersonListElement element = new PersonListElement(p);
    boolean inserted = false;
    if(containsElement(p) == true){
        inserted = false;
        return inserted;
    }
    if(head == null){
        head = element;
        size = size + 1;
        inserted = true;
        return inserted;
    }
    PersonListElement position = head;
    PersonListElement prev = null;
    while(position != null){
        if(position.getValue().compareTo(element.getValue()) < 0){
            if(prev != null){
                prev.setNext(element);
                element.setNext(position);
                inserted = true;
                position = null;
            }
            prev = position;
            position = position.getNext(); //here i get pointerException prob.
        }
    }
    if(inserted == false && prev != null){
        prev.setNext(element);
        inserted = true;
    }
    size = size + 1;
    return inserted;
}
通过这个类,我测试了我的代码:

class SortList
{
    public static void main(String args[]){
        SortedList list = new SortedList();
        Person a = new Person("Alice", 10);
        list.insert(a);
        a = new Person("Al", 20);
        list.insert(a);
        a = new Person("Bo", 90);
        list.insert(a);
        p = new Person("Cin", 3);
        list.insert(a);     
        System.out.println(list);
        }
}

您需要完成迭代和插入。假设您正在实现一个链表,并且在
liselement
类中有一个public
next
属性

public boolean insert(Person p){
        ListElement element = new StackElement(p);
        boolean inserted = false;
        if(head == null){
            head = element;
            size = size + 1;
            inserted = true;
            return inserted;
        }
        ListElement temp = head;
        ListElement prev = null;
        while(temp != null){
            if(temp.compareTo(element) < 0) {
                    if(prev != null) prev.next = element; //Keep the chain connected
                    element.next = temp; //Insert you element before temp
                    inserted = true;
                    break; 
                }
                prev = temp;
                temp = temp.next;
            }
        }
        if(!inserted && temp == null && prev != null) { //You may want to insert it at the end of the list if no upper element is found.
                 prev.next = element; //Put it at the end.
                 inserted = true;
        }
       if(inserted) size = size + 1;
       return inserted; //This should be true in any case but I left it for the case you want to avoid inserting the element at the end if its place is not found.
    }
关于你的
delete
方法,我认为你把事情弄得太复杂了。 在迭代中,您只需要保留两个引用:一个用于前一个元素,另一个用于当前元素。 如果您的第一个元素是要删除的元素,则您的
prev
引用将为
null
,并且您必须更新
。否则,您需要更新
prev
next reference,使其引用当前
之后的元素:

public boolean delete(Person p){
    boolean deleted = false;
    PersonListElement current = head;
    PersonListElement prev = null;
    while(current != null && !deleted){
        if(current.getValue().compareTo(p) == 0){
            if(prev != null) 
                prev.setNext(current.getNext());
            else
                head = current.getNext();
            deleted = true;
        } else {
        prev = current;
        current = current.getNext();
       }
    }
    if(deleted) size--;
    return deleted;
}

你的类的其余部分看起来是什么样子的?不确定你的赋值是否允许,但是Java有一些很好的排序功能,使用你的compareTo方法,你已经完成了一半。我无法理解下面的代码:if(prev!=null)prev.next=element//保持链条连接元件。下一步=温度//在temp inserted=true之前插入元素;temp=null;}prev=温度;温度=下一个温度;您能帮助我吗?@usersuper
prev
temp
之前保留对节点的引用。您要询问的
if
语句用于检查是否存在前一个元素,在这种情况下,您必须将元素放在它的正后方(通过使其属性
next
引用
element
),并且在任何情况下都要放在
temp
之前(通过制作
元素。下一步
参考
temp
).现在更清楚了吗?我接受了你的答案。我还有一个关于删除元素的问题。我试图做到这一点,但我认为如果元素位于列表的末尾,我的代码就不起作用了。我的代码如下。我删除了我的答案。如果我想删除最后一个元素,我不确定你的删除方法是否有效,因为if(prev!=null){prev.setNext(current.getNext());}。那么我认为这可能有问题,因为current.getNext()不存在
public void insert(Person p){
        ListElement element = new StackElement(p);

        if(head == null){
            head = element;
            size = size + 1;
            return;
        }

        ListElement temp = head;
        ListElement prev = null;

        boolean inserted = false;
        while(temp != null){
            if(temp.compareTo(element) < 0) {
                    if(prev != null) prev.next = element; //Keep the chain connected
                    element.next = temp; //Insert you element before temp
                    inserted = true;
                    temp = null;
                } else {
                    prev = temp;
                    temp = temp.next;
                }
            }
        }
        if(!inserted && prev != null)
        { 
                 prev.next = element; //Put it at the end.
        }
       size = size + 1;
    }
public boolean delete(Person p){
    boolean deleted = false;
    PersonListElement current = head;
    PersonListElement prev = null;
    while(current != null && !deleted){
        if(current.getValue().compareTo(p) == 0){
            if(prev != null) 
                prev.setNext(current.getNext());
            else
                head = current.getNext();
            deleted = true;
        } else {
        prev = current;
        current = current.getNext();
       }
    }
    if(deleted) size--;
    return deleted;
}