Java單鏈接列表娛樂

Question

所以對於一個任務，我需要基本上做一個單向鏈表，使用2種方法，add(x)這會在列表末尾添加一個新節點，並且會從末尾添加deleteMin()該清單

底部是我所做的代碼。我不斷收到錯誤java.lang.NullPointerException在66行

裏面加（x）的方法在head.next = u;

我已經嘗試了幾次修改代碼來修復這個錯誤，似乎沒有任何工作。

package assignment1; 

import java.util.InputMismatchException; 
import java.util.Scanner; 

public class SLL { 
    private int item; 
    private SLL next; 

    SLL(int x, SLL y){ 
     item = x; 
     next = y; 
    } 

    public static SLL head; 
    public static SLL tail; 
    public static int n; 

    public static void main(String[] args){ 
     boolean x = true; 
     Scanner user_input = new Scanner(System.in); 
     System.out.println("A = add element, R = remove element, L = to list items inside Singly-Linked-List, Q = to stop session"); 
     while (x == true) { 
      String user = user_input.next(); 
      String userLow = user.toLowerCase(); 
      switch (userLow) { 
       case "a": 
        System.out.println("Add an integer"); 
        int a; 
        try { 
         a = user_input.nextInt(); 
        } catch (InputMismatchException e) { 
         System.out.println("Incorect input, please input an integer."); 
         break; 
        } 
        System.out.println(""); 
        add(a); 
        break; 
       case "r": 
        deleteMin(); 
        break; 
       case "l": 
        //list(); 
        break; 
       case "q": 
        x = false; 
        break; 
       case "help": 
        System.out.println("A = add element, R = remove element, L = to list items inside queue, Q = to stop session"); 
        break; 
       case "?": 
        System.out.println("A = add element, R = remove element, L = to list items inside queue, Q = to stop session"); 
        break; 
       default: 
        System.out.println("Not a recognized command: try 'help' or '?' for a list of commands"); 
        break; 
      } 
     } 
    } 

    public static void add(int x){ 
     SLL u = new SLL(x, head); 
     if (n == 0) { 
      head.next = u; // <<<------------ 
      head.next.next = tail; 
     } else { 
      SLL current = head; 
      for (int i = 0; i < n; i++) { 
       current = current.next; 
      } 
      current.next = u; 
      current.next.next = tail; 
     } 
     n++; 
    } 
    public static void deleteMin(){ 
     if (n == 0) { 
      System.out.println("No elements inside list"); 
     } else if (n == 1) { 
      head.next = null; 
      tail.next = null; 
      n--; 
     } else { 
      head.next = head.next.next; 
      head.next.next = null; 
      n--; 
     } 
    } 
} 

Answer 1

您還沒有初始化head還，但嘗試訪問它的next屬性。這導致NullPointerException，因爲當時head是null。

您需要初始化您的head變量與某事物。在嘗試訪問它的值之前，像head = new SSL(item, next)一樣。

當您在該行中設置下一個屬性，並且您的SSL構造函數也會設置該屬性，因此您可能想用head = new SSL(x, u);替換head.next = u;。

你這樣做的方式，下一行也將導致NullPointerException，因爲當你指定head.next.next = tail;的head.next被null。當你將它交給SLL u = new SLL(x, head);的構造函數時，這來自head，它是null。
無論如何，此線路可能無效，因爲此時tail也是null。所以你基本上做head.next.next = null;。