如何从Java工作中的单链表中删除尾部

Question

I am reading algorithm to delete last element of a singly Linked List. Assume I have a Linked List Object called ListNode:

public class ListNode {
    private int data;
    private ListNode next;

    public ListNode(int data) {
        this.data = data;
    }

    public int getData() {
        return this.data;
    }

    public void setData(int data) {
        this.data = data;
    }

    public ListNode getNext() {
        return this.next;
    }

    public void setNext(ListNode next) {
        this.next = next;
    }
}

I found the method to delete the last node of the list is:

public ListNode deleteAtTail(ListNode head) {

    if (head == null || head.next == null) return null;
    ListNode node = head;
    while(node.next.next != null) {
        node = node.next;
    }

    node.next = null;
    return head;
}

I am confusing how this code is working since everything is through "node". However, when return head, the last node is deleted. Therefore, I wonder how it is working, is it related to "passed by value" in Java?

Answer 1

You iterate over the node s of the list until node.next.next is null. At this point, node refers to the next to last node, and node.next refers to the last node. Setting node.next to null removes the last node from the list, since no node in the list refers to it anymore.

Answer 2

You can notice that the method is iterating through all nodes till second last node because the next of last node will be null .

while(node.next.next != null) {
    node = node.next;
}

Above code will give you second last node and its next is set to null using node.next = null; This means second last node will become last node now.

Answer 3

Since none of the answers are clear (nor correct in my opinion, because if we have node.next.next and we have only 1 element, we will get a NullPointerException), I'd like to give my two cents.

There are 3 scenarios:

1. List is empty. Straightforward, return null, or print that list is empty.
2. List has 1 item. We can't know this without counting (or can we?), but look at the code below
3. List has more than one item. Go through the list and have a temp variable for the previous, then set previous.next as null when you get to the end.

So, my approach is to have an initial previous variable set as null (imagine you are before the beginning of the list). Then in the while loop, if it has 1 item, it wont execute any commands (it will skip it and prev will be null), else, go with point 3. Here's the code:

    if(head == null) return;
    ListNode iterator = head;
    ListNode prev = null;
    while(iterator.next !=null) {
        prev = iterator;
        iterator=iterator.next;
    }
    if(prev == null) head = null;
    else prev.next = null;

Hope this helps.