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// Use Node Represents a node 
public class Node {
    public int val;
    public Node next=null;

    public Node(int val) {
        this.val = val;
    }

    @Override
    public String toString() {
        return "["+val+"]";
    }
}

2. Create a linked list of fixed content

// In this way , Create a linked list of fixed contents 
    public static Node creatList(){
        // Create a node 
        Node a=new Node(1);
        Node b=new Node(2);
        Node c=new Node(3);
        Node d=new Node(4);

        // Assign values to references 
        a.next=b;// hold  b The address of is assigned to a Of next in 
        b.next=c;// hold  c The address of is assigned to b Of next in 
        c.next=d;// hold  d The address of is assigned to c Of next in 
        d.next=null;// This line can be left blank ,Node Class has been initialized to null
        return a;// Return to header node 
    }

3. Get the header

Node head=creatList();

Two 、 Traversing the linked list

Method 1 （for loop ）

for (Node cur=head;cur!=null;cur=cur.next){
            System.out.println(cur.val);
        }

Method 2 （while loop ）

 Node cur=head;
        while(cur!=null && cur.next!=null){
            cur=cur.next;
            System.out.println(cur.val);
        }

Get the length of the linked list

Just add count++ that will do .

3、 ... and 、 Find nodes as required

1. Find out whether there is a node on the linked list

        int tofind=3;// Suppose the value to be found is 3
        Node cur=head;
        for (;cur!=null;cur=cur.next){ // Traverse 
            if (cur.val==tofind){     // Judge 
                break;
            }
        }
        if (cur!=null){
            System.out.println(" eureka ");
        }else{
            System.out.println(" Did not find ");
        }

2. Take the last node of the linked list

        Node cur=head;
        while(cur!=null && cur.next!=null){
            cur=cur.next;
        }
        // Once the cycle is over ,cur It points to the last node of the linked list 
        System.out.println(cur.val);

3. Take the... Of the linked list N Nodes

       int N=3;  // Suppose you find the 3 Nodes 
        Node cur=head;
        for (int i=1;i<N;i++){
            cur=cur.next;
        }
        System.out.println(cur.val);

4. Take the penultimate of the linked list N Nodes

Last but not least N A node is a positive number size+1-N Nodes

size Indicates the length of the list , Put the N Switch to size+1-N, You can find .

5. Take the penultimate node of the linked list

Characteristic is ： .next.next If it is empty, it will jump out of the loop

        Node cur=head;
        while(cur!=null && cur.next!=null && cur.next.next!=null){
            cur=cur.next;
        }
        System.out.println(cur.val);

Judge cur!=null && cur.next!=null To prevent null pointer exceptions .

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