覆盖链表中的等于

时间:2012-12-01 18:57:02

标签: java list linked-list

一直在寻找我的问题的解决方案。已经阅读了很多关于覆盖等于正确比较链表的节点但已经卡住了。基本上我试图搜索我的列表,找到一个等于它的节点并删除它。我找到了查看链表的方法,但是当我返回节点时,它只是乱码。我试图找到一种方法将它变成一个字符串,并将其与另一个字符串进行比较。无论如何这里是代码和我的equals方法,目前无法正常工作。

public class MagazineList {
    private MagazineNode list;
    private Object obj;

    public MagazineList(){
        list = null;
    }


    public void add(Magazine mag){
        MagazineNode node = new MagazineNode(mag);
        MagazineNode current;

        if(list==null)
            list = node;
        else{
            current = list;
            while(current.next != null)
                current = current.next;
            current.next = node;
        }   
    }
    public void insert(Magazine mag)
    {
        MagazineNode node = new MagazineNode (mag);

        // make the new first node point to the current root
        node.next=list;

        // update the root to the new first node
        list=node;
    }
    public void deleteAll(){
        if(list == null){

        }
        else{
            list = null;
        }
    }

    public void delete (Magazine mag) {
        MagazineNode current = this.list;
        MagazineNode before;

        //if is the first element
        if (current.equals(mag)) {
            this.list = current.next;
            return;     //ending the method
        }


        before = current;

        //while there are elements in the list
        while ((current = current.next) != null) {

            //if is the current element
            if (current.equals(mag)) {
                before.next = current.next;
                return;     //endind the method 
            }

            before = current;
        }

        //it isnt in the list
    }
    public boolean equals(Object other) {

        System.out.println("Here in equals" + other + this);
        // Not strictly necessary, but often a good optimization
        if (this == other)
            return true;
        else{
            return false;
        }
    }

    @ Override
    public String toString(){
        String result = " ";

        MagazineNode current = list;
        while (current != null){
            result += current.magazine + "\n";
            current = current.next;     
        }
        return result;
    }


    private class MagazineNode {
        public Magazine magazine;
        public MagazineNode next;


        public MagazineNode(Magazine mag){
            magazine = mag;
            next = null;
        }
    }
}

3 个答案:

答案 0 :(得分:5)

您的删除方法看起来不错,但您不应将MagazineNode与杂志进行比较。您应该将杂志与杂志进行比较。

if (current.equals(mag))替换为if (current.magazine.equals(mag))

答案 1 :(得分:1)

好的 - 请注意,你的equals方法只是重新实现== identity equals。

这意味着如果other不是同一个Magazine对象,则会失败。如果这就是你想要的那样好,但通常你想选择杂志中的属性。

因此,如果杂志有String title,那么在你的equals方法中,你会做类似的事情:

if (magazine instanceof Magazine && magazine.getTitle().equals(other.getTitle()) returnval = true;

另外,请参阅Joshua Bloch的Effective Java,以获得对此的精彩描述。每次重写equals方法时,您还希望覆盖hashCode方法。他们走到一起,他描述了原因。

希望这会有所帮助。

答案 2 :(得分:1)

如果您使用的是eclipse,请使用此功能:Eclipse - >来源 - >生成hashCode()和equals()。并研究其实施。这将有助于您了解如何编写equals()和hashCode()。祝你好运。