删除singleLinked列表中的最后一个元素

Question

我正在制作一个简单的链表，并且试图实现一种允许删除链表的最后一个节点的方法。 在某些时候，我的方法是错误的，我不确定错误在哪里以及如何解决。 这是代码！

public Nodo deleteEnd() {

    Nodo aux;
    if (head == null) {
        throw new NoSuchElementException("Element cant be deleted");

    } else {

        aux = head;

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

        size--;
    }
    return aux;
}

Answer 1

您需要将最后但并非最不重要的节点的next分配为null ：

if(head.next == null) {
    // head is last!
    head = null;
    size = 0;
} else {
    previous = head;
    current = head.next;
    // while current is not last, keep going
    while(current.next != null) {
       previous = current;
       current = current.next;
    }
    // current is now on last!
    previous.next = null;
    size--;
}

Answer 2

尝试减少一个.next ：

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

Answer 3

public Nodo deleteEnd() {
    if (head == null) {
        throw new NoSuchElementException("Element can't be deleted");
    }
    Nodo current = head;
    Nodo next = current.next;

    // Check if there is only the head element.
    if ( next == null )
    {
        size--;
        head = null;
        return current;
    }

    // Skip to the end.
    while (next.next != null) {
        current = next;
        next = next.next;
    }

    // Break the link to the next element.
    current.next = null;
    size--;
    return next;
}

Answer 4

加

aux.next = null;

在while循环之后-那么将没有对最后一个元素的引用。