java - 单链表中节点的删除

Question

while 循环只是遍历链表的头部并表示要删除它。但没有删除。我想删除中间的一个节点，但这也没有发生。

我尝试过 equals 方法，也尝试过compareTo 方法

//******************Deletion by Key***********************
//Method to delete a node in the Linked List by key
public static LinkedList deleteByKey(LinkedList list, String first, String last) {
    //Store head node
    Giftor currNode = new Giftor(null, null);
    currNode = list.head;
    Giftor prev = new Giftor(null, null);
    //prev= null;
    System.out.println("Deleting: " + first + " " + last);
    System.out.println("currNode.firstName " + currNode.firstName);

    //
    //Case 1:
    //If head node itself holds the key to be deleted
    if(currNode != null && currNode.firstName.equals(first) && currNode.lastName.equals(last)) {
        list.head = currNode.next;
    System.out.println(first + " " + last + " found and deleted");
    return list;
    }

    //
    //CASE 2:
    //If the key is somewhere other than head
    // 

    // Search for the key to be deleted, 
    // keep track of the previous node 
    // as it is needed to change currNode.next 
    while (currNode != null && (currNode.firstName.compareTo(first) == 0
             && currNode.lastName.compareTo(last) == 0)) { 
        // If currNode does not hold key 
        // continue to next node 
        prev = currNode;  
        currNode = currNode.next; 
    }

    //If the firstName and lastName were present, it should be at currNode
    //Therefore the currNode should not be null
    if(currNode != null) {
        //Since the firstName/lastName is at currNode
        //Unlink currNode from linked list
        prev.next = currNode.next;
        //Display the message
        System.out.println(currNode.firstName + " " + currNode.lastName + " found and deleted");
    }

    // 
    // CASE 3: The key is not present 
    // 

    // If key was not present in linked list 
    // currNode should be null 
    if (currNode == null) { 
        // Display the message 
        System.out.println(currNode.firstName + " " + currNode.lastName + " not found"); 
    } 

    // return the List 
    return list; 

}//end of deleteByKey

LinkedList: 
Micheal Womack
Randall Womack
Rita Evans
Trent Beck
Chris Baird
Lisa Adams
Deleting: Rita Evans
currNode.firstName Micheal
Micheal Womack found and deleted
LinkedList: 
Micheal Womack
Randall Womack
Rita Evans
Trent Beck
Chris Baird
Lisa Adams
LinkedList: 
Micheal Womack
Randall Womack
Rita Evans
Trent Beck
Chris Baird
Lisa Adams
Deleting: Rita Evans
currNode.firstName Micheal
Micheal Womack found and deleted
LinkedList: 
Micheal Womack
Randall Womack
Rita Evans
Trent Beck
Chris Baird
Lisa Adams

Answer 1

情况 2 中的 while 循环对我来说似乎是错误的。仅当当前节点的名字和姓氏 等于 first 和 last 时，它才会进行迭代。因此，如果头节点不匹配，则循环体根本不会执行。如果确实匹配，情况 1 就已经解决了这个问题。

尝试将其更改为:

while (currNode != null && (!currNode.firstName.equals(first) 
                            || !currNode.lastName.equals(last))) { 
        // If currNode does not hold key 
        // continue to next node 
        prev = currNode;  
        currNode = currNode.next; 
    }