Question

我刚刚完成了这个程序的工作并将其编译好但是在用户输入后它会中断并给我这个：

请在键盘输入0或更多值 12 4 3 2 1

Exception in thread "main" java.lang.NullPointerException
at Search.buildList(Search.java:41)
at Search.main(Search.java:10)

以下是代码：

import java.io.*; 
import java.util.*;

public class Search { 
public static void main(String argv[]) throws IOException { 

Scanner stdin = new Scanner(System.in);
System.out.println("Please input 0 or more values at keyboard");
Node head = buildList();

System.out.println("Now printing list");
printList(head);
System.out.println("\nWhat key in list are you searching for? ");
int key = stdin.nextInt();
System.out.print("Your key was ");
if (search(head, key))
System.out.println("found.");
else
System.out.println("not found.");

}

private static void printList(Node head)
{
            if (head != null)
            {
                    System.out.print(head.getItem() + " ");
                    printList(head.getNext());
            }
}

private static Node buildList() throws IOException
{
 // Post : Inserts 0 or more numerical values from keyboard into list
//          using the Scanner class and returns head of list

Scanner input = new Scanner(System.in);
Node head = null;
Node first = new Node(input.nextInt());
head.setNext(first);
while(input.hasNext())
{    
insert(first, input.nextInt());
/*
  Node curr = new Node(input.nextInt());
  Node prev = head;
  while (true)
  {
prev = prev.getNext();
if ((int)curr.getItem() < (int)prev.getItem())
{
  head.setNext(curr);
  curr.setNext(prev);
  break;
}
if (prev.getNext() == null)
{
  prev.setNext(curr);
  break;
}
  }*/
}
return first;
} 

private static Node insert(Node head, Comparable newValue)
{
Node prev, curr = head;

for (prev = null,  curr = head;
         curr != null && newValue.compareTo(curr.getItem()) > 0;
         prev = curr, curr = curr.getNext() ) {}

    Node newNode = new Node(newValue, curr);
if (prev != null)
    {
        prev.setNext(newNode);
    return head;
    }
else
    return newNode;
}

private static boolean search(Node head, Comparable key)
{
 // PRE:  head points to the front of linked list;  list may be
 //         empty or non-empty;  key is item searching for
 // POST: returns true or false regarding whether key is found in
 //         list
if (head == null){
    return false;}
else if (head.getItem().equals(key)){
    return true;}
else {
    return search(head.getNext(), key);
}

} 

}

有什么想法吗？

输出应类似于以下内容：

请在键盘上输入0或更多值

12 4 -1 5 3 0 2

现在打印列表

-1 0 2 3 4 5 12 你在找什么钥匙？ 15 找不到你的钥匙

Answer 1

Node head = null;

无论何时在null对象上调用方法，都会得到nullPointerException。这就是head.setNext(first);给你异常的原因。所以你可以做到

而不是这个

Node head = new Node();

你将避免使用NullPointerException。

根据您的要求，您应该这样做。

private static Node buildList() throws IOException
{
 // Post : Inserts 0 or more numerical values from keyboard into list
//          using the Scanner class and returns head of list

Scanner input = new Scanner(System.in);
Node head = null;
Node first = new Node(input.nextInt());
head=first; //assigning the first value to head
while(input.hasNext())
{    
insert(first, input.nextInt());
head.setNext(first);//insert the node in the list
}
return first;
}

注意：我假设 setNext（）在列表中的适当位置插入节点而不是直接在头节点的下一个位置（否则你只会得到） 2个节点，无论您插入多少个数字）

Answer 2

Node head = null;

以上一行会将{strong> head 设为object reference variable Node到nul l，现在调用任何方法在此对象上，引用变量将导致NullPointerException。

Node head = new Node();

这条线路将是更好的方法，因为会阻止 NullPointerException。

NullPointerException使用链接列表时出错

2 个答案: