按字母顺序的字符串插入Java链接列表

Question

EDIT: Sorry, here is the code where I call the insert() method. The infinite loop is gone now, but the last case in the insert (else...) still doesn't add a String in order.

public static void main(String[] args)
    {
        Frequency f = new Frequency();
        String s1 = "apple";
        String s2 = "crayon";
        String s3 = "bear";
        String s4 = "apple";
        String s5 = "";
        f.insert(s1);
        f.insert(s2);
        f.insert(s3);
        f.insert(s4);
        f.insert(s5);
        f.print();
    }

Output of test: (apple, 2)(crayon, 1)

---

I received a project in my college course which requires me to insert Strings into a Linked List in alphabetical order. I think I have the correct method, but running the program causes an infinite loop; I'm not sure why. Here is the code I have so far, I would prefer an explanation over code:

/*
 * Inserts a word into the linked list. If the word exists, increment the 
 * count by q. 
 */
public void insert(E word){
    //word is empty
    if(word.equals("")){
        return;
    }
    //list is empty
    else if(isEmpty())
    {
        Node n = new Node(word);
        first = n;
        N++;
    }
    //list already has word
    else if(contains(word))
    {
        Node cur = first;
        while(cur != null)
        {
            if(cur.key == word)
                cur.count++;
            cur = cur.next;
        }
    }
    //list only has 1 object
    else if(N == 1)
    {
        Node n = new Node(word);
        first.next = n;
        N++;
    }
    //inserting new object
    else
    {
        Node n = new Node(word);
        Node cur = first;
        while(cur != null)
        {
            if(cur.next != null)
            {
                if(cur.next.key.compareTo(word) <= 0)
                {
                    Node temp = cur.next;
                    cur.next = n;
                    n.next = cur.next;
                    return; //exit, since the word has been added
                }
                cur = cur.next;
            }
            else
                cur.next = n;
        }
        N++;
    }
}

Answer 1

As I see you want to get the count of Keys and all Keys in the right order you may use a

TreeMap<String,Integer>

with for your purpose and insert by :

treeMap.put(key, treeMap.getOrDefault(key, 0));

To your question: If your are in the last case of your question "cur" can never get null. Because you alter the cur variable only if cur.next != null. So once cur.next is null you expect the while would break after the next round. However you don't alter cur's value anymore.

    while(cur != null)
    {
        if(cur.next != null)
        {
            if(cur.next.key.compareTo(word) <= 0)
            {
                Node temp = cur.next;
                cur.next = n;
                n.next = cur.next;
                return; //exit, since the word has been added
            }
            ---> old position of cur = cur.next;
        }
        else {
            cur.next = n;
        }
        cur = cur.next; ---> move it here
    }