無法刪除鏈表中的最後一個節點

Question

我試圖從我單鏈表中刪除最後一個節點。但是我仍然無法解決代碼中的這個錯誤。我的deleteFromEnd方法不會刪除最後一個節點。在調用delete方法後，它仍然顯示我想要刪除的節點。列表的其餘部分將被刪除，但最後一個節點本身不會被刪除。你能告訴我我錯過了什麼，或者錯誤在哪裏？

的LinkedList：

package lab5; 

public class LinkedList { 

    public static void main(String argsp[]) { 
     List ob = new List(); 

     ob.addAtStart("y", 6); 
     ob.addAtStart("w", 4); 
     ob.addAtStart("z", 3); 

     ob.addAtEnd("a", 3); 
     ob.addAtEnd("b", 4); 
     ob.addAtEnd("c", 5); 

     /* 
     * ob.display(); System.out.println("Deleted first one"); 
     * ob.deleteFromStart(); 
     */ 
     ob.display(); 
     System.out.println("Deleted End one"); 
     ob.deleteFromEnd(); 
     ob.display(); 
    } 
} 

列表：

package lab5; 

public class List { 

    Node head; 

    public List() { 
     head = null; 
    } 

    public List(Node e) { 
     head = e; 
    } 

    Node oldfirst = null; 
    Node lasthead = null; 

    public void addAtStart(String name, int age) { 
     Node newObject = new Node(name, age); 
     newObject.next = head; 

     if (oldfirst == null) { 
      oldfirst = newObject; 
     } 
     head = newObject; 
     lasthead = head; 

    } 

    public void display() { 
     Node store = head; 
     while (store != null) { 
      store.display(); 
      store = store.next; 
      System.out.println(); 
     } 
    } 

    public void addAtEnd(String name, int age) { 
     Node atEndValue = new Node(name, age); 
     oldfirst.next = atEndValue; 
     oldfirst = atEndValue; 
    } 

    public void deleteFromStart() { 
     if (head.next != null) { 
      head = head.next; 
     } 
    } 

    public void deleteFromEnd() { 
     Node start = head; 
     Node prev = null; 
     while (head != null) { 
      prev = head; 
      head = head.next; 
     } 
     prev.next = null; 
     head = prev; 
    } 

    public Node search(String name) { 
     return head; 
    } 

    public boolean isEmpty() { 
     return head == null; 
    } 

    public int size() { 
     return (head.toString()).length(); 
    } 
} 

節點：

package lab5; 

public class Node { 

    String name; 
    int age; 
    Node next; 

    public Node() { 
     name = "Abc"; 
     age = 10; 
     next = null; 
    } 

    public Node(String name, int age) { 
     this.name = name; 
     this.age = age; 
     next = null; 
    } 

    public void display() { 
     System.out.println("Name: " + name + " Age: " + age); 
    } 
} 

Answer 1

要修改名單這是不對的頭指針。以下方法爲我工作。

public void deleteFromEnd() { 
     Node start = head; 
     Node prev = null; 

     if(start == null || start.next == null) 
     { 
      head = null; 
      return; 
     } 
     while (start.next != null) { 
      prev = start; 
      start = start.next; 
     } 
     prev.next = null; 
    } 

在分析了您的代碼後，我發現了其他一些問題。您需要更新addAtStart和addAtEnd方法。

Node lasthead = null; 

public void addAtStart(String name, int age) { 
    Node newObject = new Node(name, age); 
    newObject.next = head; 
    if(head == null) 
     lasthead = newObject; 
    else if(head.next == null) 
     lasthead = head; 

    head = newObject; 

} 


public void addAtEnd(String name, int age) { 
    Node atEndValue = new Node(name, age); 
    lasthead.next = atEndValue; 
    lasthead = atEndValue; 
} 

原因是，假設如果我從列表的末尾刪除單個節點。我無法將元素添加到列表的末尾。

Answer 2

不要更改連接的L-的頭ist，否則你會失去名單。 嘗試下面的功能修改：

public void deleteFromEnd() { 
     Node start = head; 
     Node prev = null; 
     if(start == null){ 
      return; 
     } 
     if (start.next == null){ 
      head = null; 
      return; 
     } 
     while (start.next != null) { 
      prev = start; 
      start = start.next; 
     } 
     prev.next = null; 
    } 

Answer 3

當你從一個單鏈接列表的末尾刪除你要做的事情：

1. 遍歷列表，並創建一個變量來引用秒到最後你的名單中的元素。

2. 設置第二到最後一個節點後，該節點爲空

你應該永遠不會改變的head價值，同時遍歷你的鏈接列表，因爲有效地刪除整個列表。自從您覆蓋了您的head變量後，您無法找到自己的方式。相反，使用初始化爲head的臨時變量進行迭代。

最後，別忘了考慮邊緣情況下，該列表只有1元，或者已經是空的：

public void deleteFromEnd() { 
    Node current = head; 
    Node previous = null; 

    while (current != null && current.next != null) { 
     previous = current; 
     current = current.next; 
    } 

    if (current == head) { 
     head = null; 
    } 

    if (previous != null) { 
     previous.next = null; 
    } 
}