按升序将元素插入多链表

时间:2016-06-09 15:20:03

标签: java algorithm sorting data-structures linked-list

我构建了一个由Friend个对象组成的多链表。 Friend班级有nameage字段。我试图通过比较名称来按升序插入Friend个对象,但我遇到了处理两个匹配名称的代码部分的问题。

我要么得到空指针异常,要么打印列表乱序。

填充列表

public static void main(String[] args) {
    LinkedList l = new LinkedList();
    l.add("Travis", 19);
    l.add("Kyler", 14);
    l.add("Abby", 10);
    l.add("Bob", 19);
    l.add("Travis", 12);
    l.add("Zander", 99);
    l.printList();
}

LinkedList上课:

public class LinkedList {

    Friend head;
    int listcount;

    public LinkedList(){
        head = null;
        listcount = 0;
    }

    public void add(String name, int age){
        Friend node = new Friend(name,age);
        if(head==null){
            head = node;
            return;
        }
        if(head.name.compareTo(node.name) > 0){
            node.nextName = head;
            head = node;
            return;
        }
        Friend current = head;
        Friend previous = null;

        while(current.name.compareTo(node.name) < 0 && current.nextName != null){
            previous = current;
            current = current.nextName;
        }
        if(current.name.compareTo(node.name) == 0 && current.age < node.age){
            node.nextName = current.nextName;
            current.nextName = node;
        }
        previous.nextName = node;
        node.nextName = current;
    }
    public void printList(){
        Friend temp = head;
        while(temp!=null){
            temp.print();
            temp = temp.nextName;
        }
    }
}

Friend上课:

public class Friend {
    String name;
    int age;
    Friend nextName;
    Friend nextAge;

    public Friend(String name, int age){
        this.name = name;
        this.age = age;
        nextName = null;
        nextAge = null;
    }

    public void print(){
        System.out.println(name+" "+age+". ");
    }
}

1 个答案:

答案 0 :(得分:1)

请像这样更改你的add()(我的意思是你的add()方法的最后一行);

    if (current.name.compareTo(node.name) == 0 && current.age < node.age) {
        node.nextName = current.nextName;
        current.nextName = node;
    } else if (current.name.compareTo(node.name) < 0 ) {
        previous.nextName = current;            
        current.nextName = node;
    } else {
        previous.nextName = node;
        node.nextName = current;
    }

输出

  
    

Abby 10。     鲍勃19。     凯勒14。     特拉维斯12。     特拉维斯19。     赞德99。