question:

write a method max() that takes a reference to the first node in a linked list as argument and returns the value of the maxnimun key in the list. assume that all keys are positive integers, and return 0 if the list is empty

answer:

import edu.princeton.cs.algs4.*;

public class Linklist
{
    
 
private static class Node//節點
    {
        int item;
        Node next = null;
    }
    
     public Node create(int n)//創建鏈表
    {
        Node head = null;
        Node p1 = head;
        Node p2;
        for(int i = 0; i < n; i++)
        {
            p2 = new Node();
            StdOut.println(
 
"input the number of node " + i + ":");
            p2.item = StdIn.readInt();
            if(head == null)
            {
               head = p1 = p2;
            }
            else
            {
               p1.next = p2;
               p1 = p2;
            }
        }
        if(head != null)
        {
           p1.next 
 
= null;
        }
        return head;
    }
   
    public void show(Node head)//遍歷鏈表
    {
        Node current = head;
        while(current != null)
        {
            StdOut.print(current.item + "\t");
            current= current.next;
        }
        StdOut.print("\n");
    }
    
    public int max(Node head)
    {
        if(head == null)
        {
            return 0;
        }
        Node current = head;
        int max_num;
        max_num = current.item;
        while(current != null)
        {
            if(current.item > max_num)
            {
                max_num = current.item;
            }
            current = current.next;
        }
        return max_num;
    }
    
    public static void main(String[] args)
    {
        Node head;
        int n;
        Linklist linklist = new Linklist();
        StdOut.println("input the length of linklist:(>0)");
        n = StdIn.readInt();
        head = linklist.create(n);
        linklist.show(head);
        int max_num = linklist.max(head);
        StdOut.println("the max number is " + max_num);
    }
    
}

1.3.27